The enthalpy change when 2.50 kg of Glauber's salt solidifies is 605.28 kJ.
To calculate the enthalpy change when 2.50 kg of Glauber's salt (sodium sulfate decahydrate, Na2SO4·10H2O) solidifies, you can follow these steps:
1. Convert the mass of Glauber's salt to moles:
2.50 kg = 2500 g
Molar mass of Na2SO4·10H2O = (2×23) + (32) + (4×16) + (10×(2+16)) = 46 + 32 + 64 + 180 = 322 g/mol
Moles of Glauber's salt = 2500 g / 322 g/mol = 7.76 mol
2. Multiply the moles by the energy released per mole:
Energy released = 7.76 mol × 78.0 kJ/mol = 605.28 kJ
The enthalpy change when 2.50 kg of Glauber's salt solidifies is 605.28 kJ.
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A solution consisting of 11. 4 g NH4Cl in 150 ml of water is titrated with 0. 20 M KOH.
a. How many milliliters of KOH are required to reach the equivalence point?
b. Calculate {Cl-], [K+], and [NH3] at the equivalence point. Assume volumes are additive
The balanced chemical equation for the reaction between NH4Cl and KOH is:
NH4Cl + KOH → NH3 + KCl + H2O
a. To calculate the volume of 0.20 M KOH required to reach the equivalence point, we need to know the amount of NH4Cl in the solution. The amount of NH4Cl can be calculated as follows:
amount of NH4Cl = (mass of NH4Cl) / (molar mass of NH4Cl)
= 11.4 g / 53.49 g/mol
= 0.2131 mol
At the equivalence point, all of the NH4Cl has reacted with the KOH, and the number of moles of KOH added is equal to the number of moles of NH4Cl in the solution. Therefore, we can calculate the volume of KOH required as follows:
moles of KOH = moles of NH4Cl
= 0.2131 mol
volume of KOH = moles of KOH / Molarity of KOH
= 0.2131 mol / 0.20 mol/L
= 1.065 L = 1065 mL
Therefore, 1065 mL of 0.20 M KOH are required to reach the equivalence point.
b. At the equivalence point, all of the NH4Cl has been converted to NH3, K+ and Cl-. Therefore, the concentration of K+ and Cl- will be determined by the amount of KOH added, while the concentration of NH3 will be determined by the amount of NH4Cl initially present. Assuming volumes are additive, the volume of the solution at the equivalence point is 150 mL + 1065 mL = 1215 mL.
The number of moles of K+ and Cl- at the equivalence point can be calculated as follows:
moles of K+ = concentration of KOH × volume of KOH added
= 0.20 mol/L × 1.065 L
= 0.213 mol
moles of Cl- = moles of NH4Cl initially present
= 0.2131 mol
The concentration of K+ and Cl- at the equivalence point can be calculated by dividing the number of moles by the volume of the solution:
[K+] = moles of K+ / volume of solution
= 0.213 mol / 1.215 L
= 0.175 M
[Cl-] = moles of Cl- / volume of solution
= 0.2131 mol / 1.215 L
= 0.175 M
The concentration of NH3 at the equivalence point can be calculated from the amount of NH4Cl initially present, since all of the NH4Cl has been converted to NH3:
moles of NH3 = moles of NH4Cl initially present
= 0.2131 mol
The concentration of NH3 can be calculated by dividing the number of moles by the volume of the solution:
[NH3] = moles of NH3 / volume of solution
= 0.2131 mol / 1.215 L
= 0.175 M
Therefore, at the equivalence point, [Cl-] = [K+] = 0.175 M, and [NH3] = 0.175 M.
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How is entropy related to the spontaneity of a reaction?
O A. AS<0 contributes to spontaneity.
O B. AS does not affect spontaneity.
C. AS = 0 contributes to spontaneity.
D. AS> O contributes to spontaneity.
ΔS> O contributes to spontaneity. This is the relationship between entropy and spontaneity. Therefore, the correct option is option D.
Entropy is a measureable physical characteristic and a scientific notion that is frequently connected to a condition of disorder, unpredictability, or uncertainty. From classical thermodynamics, where it was originally recognised, through the microscopic description of nature in statistical physics, to the fundamentals of information theory, the phrase and concept are employed in a variety of disciplines. It has numerous applications in physics and chemistry, biological systems and how they relate to life, cosmology, economics, sociology, weather science, and information systems, especially the exchange of information. ΔS> O contributes to spontaneity.
Therefore, the correct option is option D.
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A 58. 3g sample of NH3 is reacted with 126g O2, according to this reaction what is the limiting reagent? 4NH3 + 7O2 --> 4NO + 6H2O
The ratio of NH₃ to O₂ is less than 4:7, it means that NH₃ is the limiting reagent. Therefore, NH₃ will be completely consumed before O₂ and the amount of product formed will be determined by the amount of NH₃ available.
To determine the limiting reagent, we need to compare the amount of each reactant with their respective stoichiometric coefficients in the balanced chemical equation.
First, we convert the given masses of NH₃ and O₂ to moles using their molar masses:
58.3 g NH₃ × (1 mol NH₃ ÷ 17.03 g NH₃) = 3.42 mol NH₃
126 g O₂ × (1 mol O₂ ÷ 32 g O₂) = 3.94 mol O₂
Next, we compare the number of moles of NH₃ and O₂ to the stoichiometric coefficients in the balanced equation:
NH₃ : O₂ ratio = 4:7
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How many grams of iron are produced from 300. moles of carbon monoxide reacting with 15,000. grams of ferric oxide? 3CO + Fe2O3 →2Fe + 3C02
11,169 grams of iron is produced from 300 moles of carbon monoxide reacting with 15,000 grams of ferric oxide.
The balanced chemical equation shows that 3 moles of CO react with 1 mole of [tex]Fe_2O_3[/tex] to produce 2 moles of Fe. Therefore, we can calculate the number of moles of Fe produced from 300 moles of CO reacting with [tex]Fe_2O_3[/tex] as follows:
1 mole [tex]Fe_2O_3[/tex] produces 2 moles Fe
300 moles CO produces (2/3) x 300 = 200 moles Fe (by stoichiometry)
Next, we can use the molar mass of Fe to convert moles to grams:
1 mole Fe = 55.845 g Fe
200 moles Fe = 200 x 55.845 = 11,169 g Fe
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Which one? Please help I don't understand
Based on the rate law, the equivalent expression to d[NO₂]/dt is -2k[O₃][NO₂]; option B.
What is the rate law of a chemical reaction?A rate law gives a mathematical explanation of how variations in a substance's amount affect the rate of a chemical reaction.
To determine the equivalent expression to d[NO₂]/dt, differentiate the rate law with respect to [NO₂].
d/dt[k[O₃][NO₂]] = k[d[O₃]/dt][NO₂] + k[O₃][d[NO₂]/dt]
We assume d[O₃]/dt is a constant = k1 (since it is not given in the rate law)
The coefficient for NO₂ is -2,
Substituting in the equation above:
d[NO₂]/dt = (-2k/k1)[O₃][NO₂]
d[NO₂]/dt = -2k[O₃][NO₂]/k1
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Assuming pressure is constant. There are 12. 75 mL of chemical product associated with a temperature reading of 68 degrees Celsius. What will the final temperature be if the volume increased to 5. 25 mL
The final temperature will be approximately -55.6 degrees Celsius when the volume is reduced to 5.25 mL.
According to Charles' Law, when pressure is constant, the volume of a gas is directly proportional to its temperature (in Kelvin). The formula for Charles' Law is V1/T1 = V2/T2.
First, convert the initial temperature from Celsius to Kelvin (68 + 273.15 = 341.15 K). Then, plug in the values: (12.75 mL / 341.15 K) = (5.25 mL / T2).
To solve for T2, multiply both sides by T2 and divide by 5.25 mL: T2 = (341.15 K * 5.25 mL) / 12.75 mL ≈ 139.6 K. Finally, convert back to Celsius: 139.6 K - 273.15 ≈ -55.6 degrees Celsius.
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Use the VSEPR Theory to predict the molecular geometry of the following molecules:
(Remember, you may need to draw the lewis structure before making a prediction. )
HI
CBr4
CH2Cl2
SF2
PCl3
To predict the molecular geometry of these molecules using the VSEPR theory, we first need to draw the Lewis structure for each molecule:
1. HI
Lewis structure: H-I (single bond)
The central atom (Iodine) has 7 valence electrons, and each hydrogen atom contributes 1 valence electron. Therefore, the total valence electrons in the molecule is 9.
Steric number = number of lone pairs of electrons + number of atoms bonded to central atom = 0 + 1 = 1
Molecular geometry: linear
2. CBr4
Lewis structure:
:Br-C-Br:
: | :
:Br-C-Br:
The central atom (Carbon) has 4 valence electrons, and each Bromine atom contributes 7 valence electrons. Therefore, the total valence electrons in the molecule is 32.
Steric number = number of lone pairs of electrons + number of atoms bonded to central atom = 0 + 4 = 4
Molecular geometry: tetrahedral
3. CH2Cl2
Lewis structure:
H : Cl
| :
H-C-H
| :
Cl:
The central atom (Carbon) has 4 valence electrons, each hydrogen atom contributes 1 valence electron, and each chlorine atom contributes 7 valence electrons. Therefore, the total valence electrons in the molecule is 20.
Steric number = number of lone pairs of electrons + number of atoms bonded to central atom = 2 + 4 = 6
Molecular geometry: octahedral
However, the two lone pairs of electrons on the central atom will repel the bonded pairs more than the bonded pairs will repel each other. Therefore, the shape will be bent or V-shaped.
4. SF2
Lewis structure:
F : S : F
\ /
F
The central atom (Sulfur) has 6 valence electrons, each Fluorine atom contributes 7 valence electrons. Therefore, the total valence electrons in the molecule is 20.
Steric number = number of lone pairs of electrons + number of atoms bonded to central atom = 1 + 2 = 3
Molecular geometry: trigonal planar
However, the lone pair of electrons on the central atom will repel the bonded pairs more than the bonded pairs will repel each other. Therefore, the shape will be bent or V-shaped.
5. PCl3
Lewis structure:
Cl : P : Cl
:
Cl
The central atom (Phosphorus) has 5 valence electrons, each Chlorine atom contributes 7 valence electrons. Therefore, the total valence electrons in the molecule is 26.
Steric number = number of lone pairs of electrons + number of atoms bonded to central atom = 0 + 3 = 3
Molecular geometry: trigonal planar
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During a period of discharge of a lead-acid battery, 378 grams of Pb from the anode is converted into PbSO (s). What mass of PbO,(s) in grams is reduced at the cathode during this same period?
During the discharge of a lead-acid battery, the oxidation reaction occurs at the anode where lead (Pb) is converted into lead sulfate (PbSO4) and electrons are released:
Pb(s) → PbSO4(s) + 2e-
Meanwhile, reduction occurs at the cathode where lead dioxide (PbO2) is reduced to lead sulfate (PbSO4) by gaining those electrons released at the anode:
PbO2(s) + 4H+(aq) + 2e- → PbSO4(s) + 2H2O(l)
The balanced chemical equation shows that for every two electrons transferred at the anode, one molecule of PbSO4 is formed. Therefore, the 378 grams of Pb from the anode would produce 378/207 = 1.82 moles of PbSO4.
Since the reaction at the cathode involves the reduction of PbO2 to PbSO4, the same number of moles of PbSO4 should be formed at the cathode. The molar mass of PbO2 is 239.2 g/mol, so the mass of PbO2 that is reduced at the cathode would be:
1.82 moles x 239.2 g/mol = 435.8 g
Therefore, during the same period of discharge, 435.8 grams of PbO2 would be reduced at the cathode.
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Chemistry Calculating moles.
The molarity of an aqueous solution containing 6.7 moles of potassium chloride in 0.63L is 10.6 M.
The quantity of solute molecules per litre of solution is known as a solution's molarity. In this issue, we are given the number of moles of solute (potassium chloride) and the volume of the solution. We can use this information to calculate the molarity of the solution using the following formula:
Volume of solution (V) / moles of solute (n) equals molarity (M).
Substituting the given values, we get:
Molarity (M) = 6.7 moles / 0.63 L = 10.6 M
Therefore, the molarity of the solution is 10.6 M, rounded to the tenths place.
It is important to note that molarity is a measure of concentration and is affected by both the amount of solute and the volume of the solution. Thus, it is important to accurately measure the volume of the solution to calculate the molarity correctly.
Furthermore, it is important to use caution when handling concentrated solutions such as this one, as they can be hazardous. Proper safety equipment and procedures should be followed when working with such solutions.
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How many liters of CO2 are produced when 32. 6 liters
of propane gas, C3H3 reacts with excess oxygen at STP?
C3Hg + 502 + 4H20 + 3C02
Please help!!!
3.75 moles CO₂ × 22.4 L/mole = 84 liters of CO₂ are produced when 32.6 liters of propane gas reacts with excess oxygen at STP.
Based on the balanced equation provided, 1 mole of propane gas (C₃H₈) reacts with 5 moles of oxygen gas (O₂) to produce 3 moles of carbon dioxide gas (CO₂) at STP (Standard Temperature and Pressure, which is 0°C and 1 atm pressure).
To determine the number of moles of propane gas (C₃H₈) in 32.6 liters, we need to use the Ideal Gas Law:
PV = nRT
where P is the pressure (1 atm), V is the volume (32.6 L), n is the number of moles, R is the ideal gas constant (0.0821 L•atm/mol•K), and T is the temperature in Kelvin (273 K at STP).
Rearranging the equation to solve for n, we get:
n = PV/RT = (1 atm)(32.6 L)/(0.0821 L•atm/mol•K)(273 K) = 1.25 moles of C₃H₈
Since 1 mole of C₃H₈ produces 3 moles of CO₂, we can use a mole ratio to determine the number of moles of CO₂ produced:
1.25 moles C₃H₈ × 3 moles CO₂/1 mole C₃H₈ = 3.75 moles CO₂
Finally, we can convert moles to volume at STP using the molar volume of a gas:
1 mole of gas = 22.4 L at STP
So, 3.75 moles CO₂ × 22.4 L/mole = 84 liters of CO₂ are produced when 32.6 liters of propane gas reacts with excess oxygen at STP.
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1.85 l of a gas is collected over water at 98.0 kpa and 22.0 °c. what is the volume of the dry gas at stp?
In this problem, we are given the volume of a gas collected over water at a certain temperature and pressure. We need to determine the volume of the dry gas at STP (standard temperature and pressure).
First, we need to understand why the presence of water is important in this problem. When a gas is collected over water, some of the water vapor dissolves in the gas, which affects the volume of the gas we measure. In order to account for this, we need to use the concept of vapor pressure.
The vapor pressure of water at 22.0°C is 2.64 kPa. This means that at 22.0°C and 98.0 kPa, the total pressure is the sum of the pressure due to the gas and the pressure due to the water vapor. We can use Dalton's Law of Partial Pressures to calculate the pressure due to the gas alone:
P_gas = P_total - P_water vapor
P_gas = 98.0 kPa - 2.64 kPa
P_gas = 95.36 kPa
Now we can use the Ideal Gas Law to calculate the volume of the dry gas at STP:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. At STP, P = 101.3 kPa and T = 273.15 K.
We can rearrange the Ideal Gas Law to solve for the volume of the dry gas:
V_dry gas = (V_collected gas * P_gas * T_STP) / (P_STP * T_collected gas)
where V_collected gas is the volume of the gas collected over water, T_collected gas is the temperature of the gas collected over water, and T_STP is the temperature at STP.
Plugging in the numbers, we get:
V_dry gas = (1.85 L * 95.36 kPa * 273.15 K) / (101.3 kPa * 295.15 K)
V_dry gas = 1.60 L
Therefore, the volume of the dry gas at STP is 1.60 L. It's important to note that the volume of the dry gas is smaller than the volume of the gas collected over water, because some of the volume was occupied by water vapor.
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Can someone answer the questions in the image?.
“Balancing equations”
Ans.1
blank 1 =1
blank 2 = 3
blank 3 = 2
Ans.2
blank 1 = 6
blank 2 = 4
blank 3 = 5
Ans.
blank 1 = 11
blank 2 = 7
blank 3 = 8
Hematite and magnetite are important ore minerals of ________ found in ________. A. Zinc, hydrothermal deposits b. Iron, banded iron formation (BIF) c. Copper, secondary enrichment deposits d. Aluminum, placer deposits
Hematite and magnetite are important ore minerals of iron found in banded iron formations (BIF), option B is correct.
Iron is one of the most abundant elements in the Earth's crust and is an essential component of many industrial and technological applications. Hematite (Fe₂O₃) and magnetite (Fe₃O₄) are two of the most important iron ore minerals, both of which are found in banded iron formations (BIFs).
BIFs are sedimentary rocks that were formed billions of years ago and consist of alternating layers of iron oxides (hematite or magnetite) and silica-rich chert. These formations were formed when the Earth's oceans contained high levels of dissolved iron, which reacted with oxygen produced by photosynthetic organisms to form iron oxide minerals, option B is correct.
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The complete question is:
Hematite and magnetite are important ore minerals of ________ found in ________.
A. Zinc, hydrothermal deposits
B. Iron, banded iron formation (BIF)
C. Copper, secondary enrichment deposits
D. Aluminum, placer deposits
A 30g piece of metal absorbs 1,200 joules of heat energy, and its
temperature changes from 25°C to 175°C. Calculate the specific capacity of
the metal. What is the likely metal?
Answer:
Niobium (Columbium)
Explanation:
Specific heat capacity has the units J/(kg °C). To find the heat capacity, all we need to do is organize the values so the units match up.
1200 J / (0.03 kg * 150°C) = 266.67 or 267 J/(kg °C)
The closest metal to a 267 heat capacity is Niobium I believe.
Help what’s the answer?
Answer:
in chemical reactions moles correspond to the number of molecules or atoms that go into reaction. It means that number that is in front of molecule or atom for example in this reaction you have one oxygen it means one mole of oxygen. 4 molecules of acid correspond to 4 moles of HCl. So the final answer would be:
4 moles of HCl
2 moles of H2O
2 moles of Cl2
Question 3 & what is the hydronium concentration for a solution with a poh = 12.04 o -1.08 m o.98 m 0.011 m p o 1.96 m question 4 a solution is made by combining 2.5 moles of hf (ka 3,5 x 19 and 3.5 mol click save and submit to save and submit chick save asters to small ans
For question 3, we can use the relationship pH + pOH = 14 to solve for the pH, which is 1.96.
Then, we can use the equation Kw = [[H₃O⁺][OH⁻] = 1.0 x 10⁻¹⁴ to solve for the hydronium concentration, which is 5.01 x 10⁻¹³ M.
For question 4, we can use the equation for the acid dissociation constant (Ka) to solve for the concentration of the conjugate base, F-. Ka = [H₃O⁺][F⁻]/[HF].
We know the concentration of HF is 2.5 moles, so we can convert this to molarity using the volume of the solution. Then, we can plug in the values we have and solve for [F-], which is 2.77 M. This solution will be acidic, as the Ka value is less than 1.
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consider 5 sequential reactions where the product of each reaction is the reactant of the next and the 5 percent yields are 80%, 90%, 65%, 76% and 30%. if you begin with 100 molecules of the first limiting reagent, what is the maximum number of product molecules you can form at the end of the final reaction? \textbf{hint:} remember that you cannot have parts of a molecule!
Starting with 100 molecules of the first limiting reagent, the maximum number of product molecules that can be formed at the end of the final reaction, given the yields of each reaction, is 11 molecules.
Let's call the starting number of molecules of the first limiting reagent "A". Then, the number of molecules of each reactant and product after each reaction can be represented as follows,
Reaction 1: A → B (80% yield)
Starting molecules of A = 100
Molecules of B produced = 80
Reaction 2: B → C (90% yield)
Starting molecules of B = 80
Molecules of C produced = 72
Reaction 3: C → D (65% yield)
Starting molecules of C = 72
Molecules of D produced = 46.8 (rounded to 47)
Reaction 4: D → E (76% yield)
Starting molecules of D = 47
Molecules of E produced = 35.72 (rounded to 36)
Reaction 5: E → F (30% yield)
Starting molecules of E = 36
Molecules of F produced = 10.8 (rounded to 11)
Therefore, the maximum number of product molecules that can be formed at the end of the final reaction is 11, rounded to the nearest whole number.
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What is the energy of a photon that emits a light of frequency 6. 42 x 1014 Hz?
A. 3. 10 x 10-19 J
B. 4. 25 x 10-19 J
C. 9. 69 x 10-19 J
D. 4. 67 x 10-19 J
The energy of a photon that emits a light of frequency 6. 42 x 1014 Hz is 4.25 x 10^-19 J.
The energy of a photon can be calculated using the equation:
E=hf,
where E is the energy of the photon, h is Planck's constant (6.626 x 10^-34 J.s), and f is the frequency of the light emitted by the photon.
Plugging in the given frequency of 6.42 x 10^14 Hz into the equation, we get
E=(6.626 x 10^-34 J.s)(6.42 x 10^14 Hz) = 4.25 x 10^-19 J.
Therefore, the correct answer is B i.e, 4.25 x 10^-19 J.
It should be emphasized that a photon's energy is directly linked to its frequency and inversely related to its wavelength. Therefore, light with higher frequency, such as blue light, contains more energy than light with lower frequency, such as red light.
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A 0.205g sample of caco3 is added to a flask with 7.50ml of 2.00 m hcl.
caco3(aq)+2hcl(aq)-cacl2(aq) + h2o(l) + co2
enough water is added to make a 125.0ml solution.a 10.00ml aliquot of this solution is taken and titred with 0.058 naoh
naoh (aq) + hcl - h2o + nacl
how many ml of naoh are used?
The volume of [tex]NaOH[/tex] used to titrate the[tex]HCl[/tex] is 5.80 mL
First, we need to find the number of moles of [tex]HCl[/tex] that reacted with the [tex]CaCO3[/tex].
2 mol [tex]HCl[/tex] react with 1 mol [tex]CaCO3[/tex]
Moles of [tex]HCl[/tex] = (7.50 mL) x (2.00 mol/L) = 0.015 mol [tex]HCl[/tex]
From the balanced equation, we see that 1 mol of [tex]CaCO3[/tex] reacts with 2 mol of [tex]HCl[/tex]. Therefore, the number of moles of [tex]CaCO3[/tex] in the original 0.205 g sample is:
Moles of[tex]CaCO3[/tex] = 0.205 g / 100.09 g/mol = 0.002049 mol [tex]CaCO3[/tex]
Since 1 mol of [tex]CaCO3[/tex] produces 1 mol of [tex]CO2[/tex], we have:
Moles of[tex]CO2[/tex]produced = 0.002049 mol [tex]CaCO3[/tex]
Now we need to calculate the concentration of [tex]CO2[/tex] in the final 125.0 mL solution.
Concentration of [tex]CO2[/tex] = Moles of [tex]CO2[/tex] produced / Volume of solution
Concentration of [tex]CO2[/tex] = 0.002049 mol / 0.125 L = 0.0164 mol/L
Finally, we can use the balanced equation for the titration reaction to calculate the number of moles of [tex]NaOH[/tex]used.
1 mol [tex]NaOH[/tex] reacts with 1 mol [tex]HCl[/tex]
Moles of [tex]NaOH[/tex] used = (0.058 L) x (0.1000 mol/L) = 0.0058 mol [tex]NaOH[/tex]
Since the volume of the aliquot is 10.00 mL or 0.0100 L, the concentration of [tex]HCl[/tex] is:
Concentration of [tex]HCl[/tex] = Moles of NaOH used / Volume of [tex]HCl[/tex]
Concentration of [tex]HCl[/tex] = 0.0058 mol / 0.0100 L = 0.580 M
Therefore, the volume of [tex]NaOH[/tex] used to titrate the [tex]HCl[/tex]is:
Volume of [tex]NaOH[/tex] = (0.580 M) x (0.0100 L) = 0.00580 L or 5.80 mL
So, the answer is 5.80 mL.
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Which of the following is equal to 2?
O A. 6+4 ÷ (2+1) × 3
O B. (6+4 ÷ 2) - 1×3
O
C. 6+ (4÷ 2) + 1 × 3
O D. (6 + 4)÷2-1×3
O D. (6 + 4)÷2-1×3
the cacuclator gives u the answer to this
What are two results of the uneven heating of Earth's surface?
A. Ocean currents
B. Earth's axis tilt
c. Global winds
D. Coriolis effect
SUBMIT
Pls tell me the answe
According to the question the two results of the uneven heating of Earth's surface are ocean currents and global winds.
What is currents?Currents are electrical energy that flows through a circuit. They are typically measured in amperes (amps), and they result from the flow of electrons through the circuit. Currents can be either direct or alternating, and they are used to power many electrical appliances and power systems. Direct currents are generated from sources such as batteries, while alternating currents are produced by generators and power plants. Currents can also be generated artificially, using devices such as transformers, or naturally, through processes such as lightning. The magnitude of currents depends on the voltage and resistance of the circuit. Currents can be used to control the operation of many electrical circuits and components, such as motors, relays, and switches. They can also be used to provide power for many electrical devices, including lights, computers, and other electronic equipment.
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A solution is 5 mM in each of the following ions:
number ion Ksp of M(OH)2
1 Mg2+ 1. 8e-11
2 Cd2+ 2. 5e-14
3 Co2+ 1. 6e-15
4 Zn2+ 4. 4e-17
5 Cu2+ 2. 2e-20
Indicate which of the metal ions would precipitate (or start to precipitate) at each of the following pH values. Indicate your answer with the number of the ion. Use 0 to indicate no precipitate. If more than one precipitate is expected, list the numbers in increasing order and separate them with commas. For example, 3,4,5 is ok but 5,4,3 is not.
pH = 6. 00: _______________? (1,2,3,4,5 list all that apply?)
pH = 8. 00: __________? (1,2,3,4,5 list all that apply?)
What is the pH to the nearest 0. 1 pH unit at which Cu(OH)2 begins to precipitate? pH = ______?
pH = 6.00: 0, 1, 2, 3, 4, 5 will not precipitate.
pH = 8.00: 0, 1, 2, 3, 4, 5 will not precipitate.
To determine the pH at which Cu(OH)₂ begins to precipitate, we need to calculate the hydroxide ion concentration at which the product of [Cu²⁺] and [OH⁻]² reaches the Ksp value of Cu(OH)₂ (2.2e⁻²⁰). At this point, Cu(OH)₂ will begin to precipitate. Thus, we have:
Ksp = [Cu²⁺][OH⁻]²2.2e⁻²⁰ = (5e⁻³ M)[OH⁻]²[OH⁻]² = 4.4e⁻¹⁷[OH⁻] = 2.1e⁻⁸ MpOH = -log[OH⁻] = -log(2.1e⁻⁸) = 7.68pH = 14 - pOH = 6.32 (rounded to the nearest 0.1 pH unit)Therefore, Cu(OH)₂ begins to precipitate at a pH of 6.3.
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Find the balance and net ionic equation for the statements below. Answer what you can.
1. Calcium + bromine —>
2. Aqueous nitric acid, HNO3, is mixed with aqueous barium chloride
3. Heptane, C7H16, reacts with oxygen
4. Chlorine gas reacts is bubbles through aqueous potassium iodide (write both the balanced and net ionic equation)
5. Zn (s) + Ca (NO3)2 (aq) —>
6. Aqueous sodium phosphate mixes with aqueous magnesium nitrate (write both the balanced and net ionic equation)
7. Aluminum metal is placed in aqueous zinc chloride
8. Iron (III) oxide breaks down
9. Li(OH) (ag) + HCI (aq) —>
(write both the balanced and net ionic equation)
10A. Solid sodium in water. Hint: Think water, H2O, as H(OH)
10B. What would happen if you bring a burning splint to the previous reaction?
A- The burning splint continues to burn.
B - The burning splint would make a "pop" sound.
C - The burning splint would go out.
Ca +Br2 ---> CaBr2
2HNO3 + BaCl2 --->Ba(NO3)2 +2HCl
C7H16 + 11O2 → 7CO2 + 8H2O
Cl2 + 2KI --->2KCl + I2
No reaction
2Na3PO4 + 3Mg(NO3)2 → Mg3(PO4)2 + 6NaNO3
2Al + 3ZnCl2 → 3Zn + 2AlCl3
Li(OH) (ag) + HCI (aq) —>LiCl + H2O
2Na + 2H2O → 2NaOH + H2
The burning splint would make a "pop" sound.
What is the balanced equation?A balanced equation is a chemical equation that has an equal number of atoms of each element on both the reactant and product sides.
In other words, a balanced equation follows the law of conservation of mass, which states that the total mass of the reactants must equal the total mass of the products in a chemical reaction.
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12. What is the weight/volume percent concentration of 100. ML of a 30. 0% (w/v) solution of
vitamin C after diluting to 200. ML?
The weight/volume percent concentration of the diluted solution is 15%.
The initial solution is a 30.0% (w/v) solution, which means that 30.0 grams of vitamin C is dissolved in 100 mL of the solution. Therefore, the amount of vitamin C in the initial solution is:
30.0% (w/v) = 30.0 g / 100 mL = 0.3 g/mL
The initial solution is then diluted to a final volume of 200 mL. Since the amount of vitamin C in the solution remains constant, we can use the following equation to calculate the final concentration:
CiVi = CfVf
where Ci and Vi are the initial concentration and volume, and Cf and Vf are the final concentration and volume.
We can rearrange the equation to solve for the final concentration:
Cf = (CiVi) / Vf
Substituting the values, we get:
Cf = (0.3 g/mL x 100 mL) / 200 mL
Cf = 0.15 g/mL
Finally, we can convert the concentration to weight/volume percent by multiplying by 100:
weight/volume percent = Cf x 100%
weight/volume percent = 0.15 g/mL x 100%
weight/volume percent = 15%
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A solution contains 55 grams of potassium iodide, KI, dissolved in 100 grams of water at 15 °C. How many more grams of KI would have to be added to make it a saturated solution?
In order to respond to this query, it is necessary to first define a saturated solution. When a solution reaches its maximal solubility, no more solute can dissolve in the solvent, creating a saturated solution.
Potassium iodide is the solute in this scenario, while water is the solvent. Potassium iodide is most soluble in water at a temperature of around 74.2 grammes per 100 grammes of water.
We must thus add 19.2 additional grammes of KI to the solution in order to make it saturated. This implies that there would be 74.2 grammes of KI in the entire solution.
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What is the ability to do work or produce heat?
Answer: Energy
Explanation:
Energy is the ability to do work or produce heat.
<
Based on the texts, both authors would most likely agree with which
statement?
Choose 1 answer:
A
B
Sculpting representations of historical figures was a short-lived
trend.
Lewis's works are varied in the subjects they depict.
The Death of Cleopatra is Lewis's most famous piece.
Lewis's portrait busts have overshadowed her other work.
Based on the texts, both authors would most likely agree that Lewis's works are varied in the subjects they depict.
Option B is correct.
What are Lewis's works?C. S. Lewis FBA has some notable works such as The Chronicles of Narnia, Mere Christianity The Allegory of Love, The Screwtape Letters, The Abolition of Man, The Space Trilogy Till We Have Faces Surprised by Joy: The Shape of My Early Life.
This statement indicates that Edmonia Lewis created works in a range of subjects, which is supported by her sculpting of both historical and contemporary figures, as well as mythological and biblical scenes.
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which of the following options correctly describe the oxidation of primary alcohols? select all that apply. multiple select question. primary alcohols require different oxidizing conditions than secondary alcohols. a carboxylic acid can be produced by oxidation of a primary alcohol. during oxidation, a primary alcohol will rearrange to produce a more substituted oxidation product. mild oxidizing conditions will result in an aldehyde product. harsher oxidizing conditions will produce a ketone from a primary alcohol.
The options that describe the oxidation of the primary alcohols is a carboxylic acid can be produced by oxidation of a primary alcohol. Mild oxidizing conditions will result in an aldehyde product.
The Primary alcohols will be oxidized to form the aldehydes and the carboxylic acids. The secondary alcohols will be oxidized to give the ketones. The Tertiary alcohols, in the contrast, cannot be oxidized by without breaking the molecules of the C–C bonds.
The Primary alcohols and the aldehydes will be normally oxidized to the carboxylic acids using the potassium dichromate solution in the presence of the dilute sulfuric acid that is H₂SO₄.
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To what pressure must a gas be compressed in order to get into a 3. 00L the entire weight of a gas that occupies 350. 0L at standard pressure?
To answer this question, we need to use the ideal gas law, which is PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature. We also need to use the concept of molar volume, which is the volume occupied by one mole of a gas at a specific temperature and pressure.
First, we need to find the number of moles of gas that occupies 350.0L at standard pressure (1 atm) and temperature (273 K). This can be calculated using the formula n = PV/RT, where P = 1 atm, V = 350.0L, R = 0.08206 L atm/mol K, and T = 273 K. Substituting these values, we get n = (1 atm x 350.0L)/(0.08206 L atm/mol K x 273 K) = 14.15 mol.
Next, we need to find the molar volume of the gas at the pressure and volume we want it to occupy. Using the same formula, but with the new pressure (P') and volume (V') values, we get V' = nRT/P'. Since we want the gas to occupy 3.00L, we have V' = 3.00L. We also know that the number of moles (n) and temperature (T) are constant, so we can rearrange the formula to solve for the new pressure (P'). Thus, P' = nRT/V' = (14.15 mol x 0.08206 L atm/mol K x 273 K)/3.00L = 2,062.58 atm.
Therefore, the gas must be compressed to a pressure of 2,062.58 atm in order to occupy a volume of 3.00L, assuming constant temperature and number of moles. This is a very high pressure, and it highlights the importance of understanding the properties of gases and how they behave under different conditions.
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There are no attractive or repulsive forces between gas molecules. How does that affect the motion of gas particles?
The absence of attractive or repulsive forces between gas molecules means that they are free to move independently and randomly. This results in the motion of gas particles being characterized by constant collisions and changes in direction and speed. Without any forces to constrain their movement, gas particles will continue to move until they collide with other particles or the walls of their container. This is what causes gases to fill up any container they are in, as their independent motion allows them to spread out evenly throughout the available space.
What is attractive force?
An attractive force is a force that pulls or draws two or more objects or particles towards each other. It is the opposite of a repulsive force, which pushes objects or particles away from each other.
Attractive forces can be observed in a variety of contexts, including gravity, electromagnetism, and intermolecular forces in chemistry. For example, the force of gravity between two objects is an attractive force that pulls them together, while the electromagnetic force between opposite charges is also an attractive force.
What is repulsive force?
A repulsive force is a force that pushes two or more objects or particles away from each other. It is the opposite of an attractive force, which pulls objects or particles towards each other.
Repulsive forces can be observed in a variety of contexts, including electromagnetism and intermolecular forces in chemistry. For example, the force between two like charges is repulsive, while the force between two like magnetic poles is also repulsive.
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