Answer:
A substance that increases H3O+ concentration when it is dissolved in water.
Explanation:
Note that H3O+ and H+ are used quite interchangeably in chemistry.
An acid makes the H+ content higher, thereby decreasing the pH.
Answer:
a
A substance that increases H3O+ concentration when it is dissolved in water.
Explanation:
Click on the Delta H changes sign whan a process is reversed button within the activity and analyze the relationship between the two reactions that are displayed. The reaction that was on the screen when you started and its derivative demonstrate that the reaction enthalpy, ΔH, changes sign when a process is reversed. Consider the reaction H2O(l)→H2O(g), ΔH =44.0kJ What will ΔH be for the reaction if it is reversed?
Answer:
ΔH = - 44.0kJ
Explanation:
H2O(l)→H2O(g), ΔH =44.0kJ
In the reaction above, liquid water changes to gaseous water. This occurs through a process known as boiling. This process requires heat, hence the ΔH is positive.
If he reaction is reversed, we have;
H2O(g)→H2O(l)
In this reaction, gaseous water changes to liquid water. This process is known as condensation. The water vapor loses heat in this reaction. Hence ΔH would be negative but still have the same value.
Arrange the following compounds in order of increasing solubility in
water and explain your sequence.
C7H15OH C6H13OH C6H6 C2H5OH
Answer:
C6H6<C7H15OH<C6H13OH<C2H5OH
Explanation:
Organic substances are ordinarily nonpolar. This means that they do not dissolve in water. However, certain homologous series of organic compounds actually dissolve in water because they possess certain functional groups that effectively interact with water via hydrogen bonding.
A typical example of this is alcohol family. All members of this homologous series contain the -OH functional group. This group can effectively interact with water via hydrogen bonding, leading to the dissolution of low molecular weight alcohols in water.
Low molecular weight alcohols are miscible with water in all proportions. This implies that they are highly soluble in water. However, as the size of the alkyl moiety in the alcohol increases, the solubility of the alcohol in water decreases due to less effective interaction of the -OH group with water via hydrogen bonding. This explains the fact that C2H5OH is the most soluble alcohol in the list.
C6H6 is insoluble in water since it is purely a hydrocarbon with no -OH group capable of interaction with water via hydrogen bonding.
Explain, in terms of charge distribution, why a molecule of carbon tetrachloride is a nonpolar molecule.
Answer:
Explanation:
The covalent bond is the chemical bond between atoms where electrons are shared, forming a molecule. Covalent bonds are established between non-metallic elements, such as hydrogen H, oxygen O and chlorine Cl. These elements have many electrons in their outermost level (valence electrons) and have a tendency to gain electrons to acquire the stability of the electronic structure of noble gas.
The covalent bond between two atoms can be polar or nonpolar. If the atoms are equal, the bond will be nonpolar (since no atom attracts electrons more strongly). But, if the atoms are different, the bond will be polarized towards the most electronegative atom, because it will be the atom that attracts the electron pair with more force. Then it will be polar.
It can occur in a molecule that the bonds are polar and the molecule is nonpolar. This occurs because of the geometry of the molecule, which causes them to cancel the different equal polar bonds of the molecule.
In carbon tetrachloride the bonds are polar, but the tetrahedral geometry of the molecule causes all four dipoles to cancel out and the molecule to be apolar.
The carbon tetrachloride s CCL4 is a carbon molecule and four chloride molecule's. The carbon tetrachloride is a nonpolar, as the dipole movement of the molecules ae evenly spaced around the central carbon atom.
As polarity is a result of the geometric sum of the bond dipoles. The bond has separate charges. It's nonpolar because the charge is symmetrically distributed.Hence the carbon tetrachloride happens to be a nonpolar molecular.
Learn more about the in terms of charge distribution, why a molecule.
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The functional groups in an organic compound can frequently be deduced from its infrared absorption spectrum. A compound containing C, H, and O exhibits broad absorption at 3450 cm^-1 (m) and an intense band at 1725, plus a band at 1100 cm^-1 (m). Relative absorption intensity: (s) = strong, (m) = medium, (w) = weak.
What functional class(es) does the compound belong to?
List only classes for which evidence is given here. Attach no significance to evidence not cited explicitly. Do not over-interpret exact absorption band positions. None of your inferences should depend on small differences like 10 to 20 cm^-1.
The functional class(es) of this compound is(are)________. (Enter letters from the table below, in any order, with no spaces or commas.)
a. alkane (List only if no other functional class applies.)
b. alkene h. amine
c. terminal alkyne i. aldehyde or ketone
d. internal alkyne j. carboxylic acid
e. arene k. ester
f. alcohol l. nitrile
g. ether
Answer:
The functional class(es) of this compound is(are):
alcoholcarboxylic acidesterExplanation:
3450 cm-1 is indicative of OH stretching
1725 cm-1 is indicative of carbonyl group C=O
1100cm-1 shows carbon is bonded to electronegative element e.g C-O
Further information on molecular formula is required for proper structural elucidation
Calculate the pH of a 0.020 M H2CO3 solution. At 25 °C, Ka1 = 4.3 × 10-7. H2CO3(aq) + H2O(l) ↔ H3O+(aq) + HCO3-(aq)
Answer:
Explanation:
H₂CO₃(aq) + H₂O(l) ↔ H₃O⁺(aq) + HCO₃⁻(aq)
Let d be the degree of dissociation
.02( 1-d ) .02d .02d
Dissociation constant Ka₁ is given
4.3 x 10⁻⁷ = .02d x .02d / .02( 1-d )
= .004 d² / .02 ( neglecting d in denominator )
= .02 d²
d² = 215 x 10⁻⁷
d = 4.636 x 10⁻³
= .004636
concentration of H₃O⁺
= d x .02
= .004636 x .02
= 9.272 x 10⁻⁵
pH = - log [ H₃O⁺ ]
= - log ( 9.272 x 10⁻⁵ )
5 - log 9.272
= 5 - .967
= 4.033 .
What are the number of protons, neutrons, and electrons in 19 F
9?
Answer:
This isotope of fluorine has 9 protons, 9 electrons and 10 neutrons.
Explanation:
In general,for a gas at a constant volume?
Answer:
The pressure of a gas is directly proportional to its Kelvin temperature if the volume is kept constant. At constant volume and temperature, the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the component gases.
Explanation:
reasons for good care on computer
answer
1)maximise your software efficiency
2)Prevention against viruses and malware
3)Early detection of problematic issues
4)prevent data loss
5)Speed up your computer
A gas company in Massachusetts charges $2.80 for 15.0 ft3 of natural gas (CH4) measured at 20.0°C and 1.00 atm. Calculate the cost of heating 2.00 × 102 mL of water (enough to make a cup of coffee or tea) from 20.0°C to 100.0°C. Assume that only 50.0% of the heat generated by the combustion is used to heat the water; the rest of the heat is lost to the surroundings. Assume that the products of the combustion of methane are CO2(g) and H2O(l).
Answer:
$0.0238
Explanation:
The energy you need to increase the temperature of water from 20°C to 100°C is obtained from:
Q = C×m×ΔT
Where Q is the energy, C is specific heat of water (4.184J/g°C), m is mass of water (2.00x10²g - Density of water 1g/mL), ΔT is change in temperature (100.0°C - 20.0°C)
Replacing:
Q = 4.184J/g°C × 2.00x10²g × 80.0°C
Q = 66944J = 66.944kJ
As you are assuming the energy of combustion will be just 50.0% to heat the water the energy you need is 66.944kJ × 2 = 133.888kJ
The combustion of methane is:
CH4(g) + 2O2(g) ⟶ CO2(g) + 2H2O(l) ΔH = −890.8kJ
That means 1 mole of methane produce 890.8kJ. As you need 133.888kJ, moles of methane are:
133.888kJ × (1 mol CH₄ / 890.8kJ) = 0.150 moles of CH₄.
Using PV = nRT, moles of 15.0ft³ (424.8L) at 20.0°C (293.15K) and 1.00atm:
1.00atmₓ424.8L = moles CH₄ₓ0.082atmL/molKₓ293.15K
17.67 = moles CH₄
As 17.67 moles of CH₄ cost $2.80, the cost of 0.150 moles of CH₄ is:
0.150 moles CH₄ ₓ ($2.80 / 17.67 moles) =
$0.0238Which of the following is a property of matter?
The major properties of matter are volume, mass, and shape.
All matter however too is made up of tiny particles known as atoms.
Other characteristics properties of matter which can be measured include object's density, color, length, malleability, melting point, hardness, odor, temperature, and others
Learn more about matter:
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The property of matter should be volume, mass, and shape.
The following information should be considered:
The matter should be made up of small & tiny particles that we called as the atoms. It involved the density of an object, length, temperature, melting point, etc.Learn more: https://brainly.com/question/1979431?referrer=searchResults
An ideal gas sealed in a rigid 4.86-L cylinder, initially at pressure Pi=10.90 atm, is cooled until the pressure in the cylinder is Pf=1.24 atm. What is the enthalpy change for this process? ΔH =
Answer:
[tex]\Delta H=-11897J[/tex]
Explanation:
Hello,
In this case, it is widely known that for isochoric processes, the change in the enthalpy is computed by:
[tex]\Delta H=\Delta U+V\Delta P[/tex]
Whereas the change in the internal energy is computed by:
[tex]\Delta U=nCv\Delta T[/tex]
So we compute the initial and final temperatures for one mole of the ideal gas:
[tex]T_1= \frac{P_1V}{nR}=\frac{10.90atm*4.86L}{0.082*n}=\frac{646.02K }{n} \\\\T_2= \frac{P_2V}{nR}=\frac{1.24atm*4.86L}{0.082*n}=\frac{73.49K }{n}[/tex]
Next, the change in the internal energy, since the volume-constant specific heat could be assumed as ³/₂R:
[tex]\Delta U=1mol*\frac{3}{2} (8.314\frac{J}{mol*K} )*(73.49K-646.02K )=-7140J[/tex]
Then, the volume-pressure product in Joules:
[tex]V\Delta P=4.86L*\frac{1m^3}{1000L} *(1.24atm-10.90atm)*\frac{101325Pa}{1atm} \\\\V\Delta P=-4756.96J[/tex]
Finally, the change in the enthalpy for the process:
[tex]\Delta H=-7140J-4757J\\\\\Delta H=-11897J[/tex]
Best regards.
The change in enthalpy is 70.42J
Data;
Volume of the gas = 4.86LInitial Pressure = 10.90 atmFinal Pressure = 1.24 atmChange in Enthalpy = ?Change in EnthalpyThe change of enthalpy is calculated as
[tex]\delta H = \delta V + \delta nRT\\\delta n = 0\\\delta H = \delta U \\[/tex]
The volume change is negligible
The change in enthalpy here is equal to change in internal energy over ΔE
[tex]\delta H = \delta U = nCv\delta T\\\delta H = \frac{3}{2}(nR\delta T)\\\delta H = \frac{3}{2}\{\delta PV)\\ \delta H = \frac{3}{2}[(10.90-1.24)*4.86] \\\delta H = 70.42J[/tex]
The change in enthalpy is 70.42J
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How many protons are in an ion with 36 electrons and a -1 charge
Answer:
Explanation:
There are 35 protons.
The number of electrons = 36 electrons gives a -1 charge.
Where did all the other minus charges go?
They must be balanced by 35 protons.
At 25 °C, what is the hydroxide ion concentration, [OH−] , in an aqueous solution with a hydrogen ion concentration of [H+]=1.9×10−9 M?
Answer:
The correct answer will be "5.26 × 10⁻⁶".
Explanation:
The given values is:
[tex][H^{+}]=1.9\times 10^{-9} M[/tex]
As we know,
⇒ [tex]pH+pOH=14[/tex]
On taking log, we get
⇒ [tex]-log[H^{+}] + -log[OH^{-}] = 14[/tex]
Now,
Taking "log" as common, we get
⇒ [tex]log[H^{+}][OH^{-}]= -14[/tex]
⇒ [tex][H^{+}][OH^{-}]= 10^{-14}[/tex]
⇒ [tex][OH^{-}]=\frac{10^{-14}}{[H^{+}]}[/tex]
On putting the estimated value of "[tex][H^{+}][/tex]", we get
⇒ [tex]=\frac{10^{-14}}{1.9\times 10^{-9}}[/tex]
⇒ [tex]=5.26\times 10^{-6}[/tex]
Water was poured over a large oil fire to extinguish it. What would happen and why?
Answer:
I think that the fire will continue burning, because the oil and water don't mix and the water is heavier (denser) than oil, so the oil will go up and the fire with it. That's why because the gas station have sand instead of water
Water is heavier than oil. Because oil is lighter and immiscible with water, it will form a separate layer above the surface of the water and continue to burn when water is poured on a large oil fire. As a result, the fire won't be put out.
What happens when you pour water on an oil fire?A small amount of water will instantly sink to the bottom of a pan or deep fryer filled with hot, burning oil and explode there. The Scientific American claims that the characteristics of oils explain why they do not mix with water.
Oil or petroleum-related fires cannot be put out with water. Water sinks below the oil because it is heavier than oil and does not float, allowing the fire to continue to burn. Oil and petroleum fires can be put out with fire extinguishers or sand.
The temperature of the burning substance is lowered by water. The fire goes out when the temperature drops below the burning substance's ignition temperature. Here, the water serves as an acclimatizer.
Thus, it will form a separate layer above the surface of the water and continue to burn when water is poured on a large oil fire.
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#SPJ6
symbol for carboxylic acid
A sample of chemical X is found to contain 5.0 grams of oxygen, 10.0 grams of carbon, and 20.0 grams of nitrogen. The law of definite proportion would predict that a 75 gram sample of chemical X should contain how many grams of carbon
Answer: 75 grams sample of chemical X should contain 21.43 grams of carbon
Explanation: The law of definite proportion states that a given chemical compound always contains its component elements in fixed ratio.
From the question, chemical X contains 5.0 grams of oxygen, 10.0 grams of carbon, and 20.0 grams of nitrogen.
Sum up the masses
5.0g + 10.0g + 20.0g = 35.0g
This means, 10.0 grams of carbon are present in 35.0 grams of chemical X.
Now, to the determine the mass of carbon that 75 gram sample of chemical X should contain,
According to the law of definite proportion, the component elements of a given chemical compound are in fixed ratio. Therefore,
If 35.0g of chemical X contains 10.0g of carbon
Then, 75 g of chemical X will contain
(75 × 10) / 35 g
= 21.43 grams
Hence, 75 grams sample of chemical X should contain 21.43 grams of carbon.
Answer:
According to the law of definite proportion, a 75 gram sample of chemical X should contain 21.249 grams of carbon.
Explanation:
The total mass of the sample is equal to the sum of masses of oxygen, carbon and nitrogen. That is:
[tex]m_{tot} = m_{O} + m_{C} + m_{N}[/tex]
If [tex]m_{O} = 5\,g[/tex], [tex]m_{C} = 10\,g[/tex] and [tex]m_{N} = 20\,g[/tex], then:
[tex]m_{tot} = 35\,g[/tex]
According to the law of definite proportion, the following simple rule of three is used:
[tex]m_{C'} = m_{C} \times \frac{m_{tot'}}{m_{tot}}[/tex]
If [tex]m_{C} = 10\,g[/tex], [tex]m_{tot} = 35\,g[/tex] and [tex]m_{tot'} = 75\,g[/tex], then:
[tex]m_{C'} = 10\,g\times \frac{75\,g}{35\,g}[/tex]
[tex]m_{C'} = 21.429\,g[/tex]
According to the law of definite proportion, a 75 gram sample of chemical X should contain 21.249 grams of carbon.
In the activity, click on the Keq and ΔG∘ quantities to observe how they are related. Calculate ΔG∘using this relationship and the equilibrium constant (Keq) obtained in Part A at T=298K:Keq=1.24×1020Express the Gibbs free energy (ΔG∘) in joules to three significant figures.
Answer: The Gibbs free energy of the reaction is -114629.4 J
Explanation:
To calculate the Gibbs free energy of the reaction, we use the equation:
[tex]\Delta G^o=-RT\ln K_{eq}[/tex]
where,
[tex]\Delta G^o[/tex] = Gibbs free energy of the reaction = ?
R = Gas constant = [tex]8.314 J/K.mol[/tex]
T = temperature of the reaction = 298 K
[tex]K_{eq}[/tex] = equilibrium constant of the reaction = [tex]1.24\times 10^{20}[/tex]
Putting values in above equation, we get:
[tex]\Delta G^o=-(8.314J/mol.K\times 298K\times \ln (1.24\times 10^{20}))\\\\\Delta G^o=-114629.4J[/tex]
Hence, the Gibbs free energy of the reaction is -114629.4 J
A student sets up the following equation to solve a problem in solution stoichiometry. (The ? stands for a number the student is going to calculate.) Enter the units of the student's answer
(1.26mol)1mL10^−3L / (7.9mol/L)= ?
Answer:
mL * L²
Explanation:
The question in t his problem is to calculate the units of the final answer.
The units in the numerator is mol, ml and L.
The unit in the denominator is mol/L
This leads us to;
Numerator / Denominator = mol * mL * L / (mol / L )
The final units is mL * L²
what is the balanced equation for calcium sulfate?
Answer:
CaSO4
Explanation:
Calcium sulfate (or calcium sulphate) is the inorganic compound with the formula CaSO4 and related hydrates.
Real images can be projected onto a screen.
A. True
B. False
Answer:
True
Explanation:
A real image can be projected or seen on a screen but a virtual image cannot because a real image is formed when light rays coming from an object actually meet at a point after refraction through a lens while a virtual image is formed when light rays coming from an object only appear to meet at a point when produced ...
Hope this helps you, and Good luck!
Consider the reaction C12H22O11 (s) + 12 O2 (g) → 12 CO2 (g) + 11 H2O (l) in which 10.8 g of sucrose, C12H22O11, was burned in a bomb calorimeter with a heat capacity of 7.50 kJ/oC (including its water). The temperature inside the calorimeter was found to increase by 20.5 oC. Based on this information, what is the heat of this reaction per mole of sucrose? Enter your answer numerically, in terms of kJ/
Answer:
THE HEAT OF THIS REACTION PER MOLE OF SUCROSE IS 4868.75 KJ OF HEAT.
Explanation:
To answer this question:
First calculate the total heat given off by sucrose:
Total energy/ heat = heat capacity * change in temperature
Heat capacity = 7.50 kJ/ °C
Change in temperature = 20.5 °C
Heat = 7.50 kJ * 20.5 °C
Heat = 153.75 kJ of heat.
Next is to calculate the heat of reaction per mole of the sucrose
Equation of the reaction:
C12H22011 (s) + 12 O2 (g) ---------> 12 CO2 (g) + 11 H20(l)
Since 1 mole of sucrose will be the molar mass of sucrose, then we should calculate the molar mass of sucrose.
Molar mass of sucrose = ( 12* 12 + 1 * 22+ 16*11) g/mol
Molar mass = 342 g/mol of sucrose
Since 10.8 g of sucrose produces 153.75 kJ of heat, 342 g will produces how many joules of heat?
10.8 g of sucrose = 153.75 kJ of heat
342 g of sucrose = ( 342 * 153.75 kJ / 10.8)
= 52 582.5 kJ / 10.8
= 4868.75 kJ of heat
So therefore, 1 mole of sucrose will produce 4868.75 kJ of heat.
What is the mass of 3.75 moles of NaCI? ( Na= 22.99g/mol, CI= 35.45 g/mol)
Answer:
219.15 grams
Explanation:
What is the mass of 3.75 moles of NaCI? ( Na= 22.99g/mol, CI= 35.45 g/mol)
Mole of Na = 22.99g
Mole of Cl = 35.45g
For NaCl we have ratio of 1:1, so we have 1 Na for every Cl
So we just add the two together to get the molar mass of NaCl which is
22.99 + 35.45 = 58.44g/mol
And we know we have 3.75 moles of NaCl so we multiply that by the molar mass of NaCl to get our answer
3.75 x 58.44 = 219.15grams
Quantum number of boron
Answer:
The answer is 5.
What is Keq for the reaction 2HCl(9) = H2(g) + Cl2(g)?
Answer:
Keq= [(Cl2) (H2)] / (HCl)^2
Explanation:
Equilibrium Constant, Keq, is written as products/reactants.
So it's going to be Keq= [(Cl2) (H2)] / (HCl)^2
When 25ml of sulphuric acid, was titrated with 0.0820 mol/L sodium hydroxide solution the end point was detected (with phenolphthalein) at 22.5ml . Calculate the concentration of sulphuric
acid in mol/L.
Answer:
the concentration of sulphuric acid is 14mol/l
complite the following reactions. NaOH(aq)+FeBr3(aq)→
Answer:
3NaOH+FeBr3>3NaBr+
Fe(OH)3
Explanation:
After writing the equation it has to be balanced
how many grams of NH3 can be produced from 2.51 mil of N2 and excess H2 ?
please help! due in a bit
Answer:
85.34g of NH3
Explanation:
Step 1:
The balanced equation for the reaction. This is given below:
N2 + 3H2 —> 2NH3
Step 2:
Determination of the number of moles of NH3 produced by the reaction of 2.51 moles of N2. This is illustrated below:
From the balanced equation above,
1 mole of N2 reacted to produce 2 moles of NH3.
Therefore, 2.51 moles of N2 will react to produce = (2.51 x 2)/1 = 5.02 moles of NH3.
Therefore, 5.02 moles of NH3 is produced from the reaction.
Step 3:
Conversion of 5.02 moles of NH3 to grams. This is illustrated below:
Molar mass of NH3 = 14 + (3x1) = 17g/mol
Number of mole of NH3 = 5.02 moles
Mass of NH3 =..?
Mass = mole x molar Mass
Mass of NH3 = 5.02 x 17
Mass of NH3 = 85.34g
Therefore, 85.34g of NH3 is produced.
Match the following aqueous solutions with the appropriate letter from the column on the right. 1. 0.13 m FeCl3 A. Highest boiling point 2. 0.19 m Mg(CH3COO)2 B. Second highest boiling point 3. 0.30 m KI C. Third highest boiling point 4. 0.53 m Glucose(nonelectrolyte) D. Lowest boiling point An error has been detected in your answer. Check for typos,
Answer:0.30 m KI ---- A. Highest boiling point
0.19 m Mg(CH3COO)2 ---- B. Second highest boiling point
0.53 m Glucose(nonelectrolyte) ---- Third highest boiling point-C
0.13 m FeCl3---- Lowest boiling point-D
Explanation:
Using the boilng point elevation formula
ΔTb=m* kb *i
where m= molality
kb= elevated boiling point constant( here kb values will be same for all soluton)
i= vant hoff factor = number of ions present in a solution
Using the number of ions and molarity present in a solution as a collagative property, since kb is constant, we can determine which of the species has the highest boiling point.
1.) 0.13 m FeCl3= Fe³⁻ + Cl⁻
i=4
ΔTb=m* kb* i= molarity x number of ionsx Kb= 0.13 x 4= 0.52kb
2) 0.19 m Mg(CH3COO)2 = Mg²⁺ + CH₃COO⁻
i= 3
ΔTb=m* kb* i= molarity x number of ions= 0.19 x 3= 0.57kb
3. 0.30 m KI = K⁺ + I⁻
i= 2
ΔTb=m *kb *= imolarity x number of ions xKb= 0.30x 2= 0.60kb
4. 0.53 m Glucose(nonelectrolyte) =
i= 1 for nonelectroytes
ΔTb=m* kb* i = molarity x number of ionsx Kb= 0.53 x 1= 0.53Kb
therefore,
0.30 m KI ---- A. Highest boiling point
0.19 m Mg(CH3COO)2 ---- B. Second highest boiling point
0.53 m Glucose(nonelectrolyte) ---- Third highest boiling point
0.13 m FeCl3---- Lowest boiling point
What is the molar mass of P2O5?
Answer:
142 grams
Explanation:
To find the molar mass of a molecule or compound, you simply need to add together the molar masses of all of the atoms that comprise it. Phosphorus has a molar mass of about 31, while oxygen has one of about 16, meaning that the molar mass of this molecule is:
2(31)+5(16)=62+80=142
Hope this helps!
Empirical formula for compound of 2.17 mol N and 4.35 mol O
Answer:
Explanation:
ratio of moles of N and O in molecule =
N / O = 2.17 / 4.35
1/2
empirical formula = NO₂
