Answers
The total mass of products obtained when 130 g of zinc react completely with HCl is 274 g (3rd option)
How do i determine the total mass of products obtained?First, we shall determine the mass of each product obtained. Details below:
For ZnCl₂
2HCl + Zn -> ZnCl₂ + H₂
Molar mass of Zn = 65 g/molMass of Zn from the balanced equation = 1 × 65 = 65 g Molar mass of ZnCl₂ = 135 g/molMass of ZnCl₂ from the balanced equation = 1 × 135 = 135 gFrom the balanced equation above,
65 g of Zn reacted to produce 135 g of ZnCl₂
Therefore,
130 g of Zn will react to produce = (130 × 135) / 65 = 270 g of ZnCl₂
Thus, the mass of ZnCl₂ obtained is 270 g
For H₂
2HCl + Zn -> ZnCl₂ + H₂
Molar mass of Zn = 65 g/molMass of Zn from the balanced equation = 1 × 65 = 65 g Molar mass of H₂ = 2 g/molMass of H₂ from the balanced equation = 1 × 2 = 2 gFrom the balanced equation above,
65 g of Zn reacted to produce 2 g of H₂
Therefore,
130 g of Zn will react to produce = (130 × 2) / 65 = 4 g of H₂
Thus, the mass of H₂ obtained is 4 g
Finally, we shall determine the total mass of the product produced. Details below:
Mass of ZnCl₂ = 270 gMass of H₂ = 4 gTotal mass of product =?Total mass of product = mass of ZnCl₂ + mass of H₂
Total mass of product = 270 + 4
Total mass of product = 274 g (3rd option)
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Related Questions
A 0.4856 g sample of solid silver oxide is heated. Find the volume of O2 that can be released at STP.
Answers
The volume of O2 that can be released at STP from the given sample of silver oxide is 23.45 mL.
To solve this problem
Creating the balanced chemical equation for the breakdown of silver oxide is the first step in tackling this issue:
2Ag2O(s) → 4Ag(s) + O2(g)
We can deduce from the equation that 2 moles of AgO will result in 1 mole of O2. Since Ag2O has a molar mass of 231.735 g/mol, 0.4856 g of Ag2O is equivalent to:
0.4856 g Ag2O x (1 mol Ag2O/231.735 g Ag2O) = 0.002095 mol Ag2O
Therefore, the number of moles of O2 that can be produced from 0.4856 g of Ag2O is:
0.002095 mol Ag2O x (1 mol O2/2 mol Ag2O) = 0.0010475 mol O2
1 mole of any gas takes up 22.4 L of space at STP As a result, 0.4856 g of Ag2O can generate the following amount of O2 at STP:
0.0010475 mol O2 x 22.4 L/mol = 0.02345 L or 23.45 mL
Therefore, the volume of O2 that can be released at STP from the given sample of silver oxide is 23.45 mL.
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Chemistry..... Reaction Rate
W → U + S Chemistry Reaction Rate use the table to find reaction rate
See reaction Rate Table Picture
Answers
The reaction rates for trial 1 is 8.22 x 10⁻² M⁻² s⁻¹ and 1.10 M⁻² s⁻¹ for trail 2 and 3
How to find reaction rate?Keep the concentration of W constant while varying the concentrations of U and S while measuring the reaction rate in order to determine the reaction rate with regard to U and S.
Select trial 1 as the reference trial and calculate the reaction's rate constant (k) with respect to U and S, assuming that the concentration of W is constant throughout all three trials.
For trial 1:
[W] = 0.13 M
Rate = 4.72 x 10⁻⁴ M/s
For trial 2:
[W] = 0.13 M
Rate = 1.18 x 10⁻² M/s
From the equation rate = k[U][S], set up the following ratio of rates:
Rate2/Rate1 = (k[U]2[S]2)/(k[U]1[S]1)
Simplifying:
k = (Rate2/Rate1) x (1/[U]2) x (1/[S]2) x ([U]1) x ([S]1)
Substituting the values from trials 1 and 2:
k = (1.18 x 10⁻² M/s) / (4.72 x 10⁻⁴ M/s) x (1/0.65 M) x (1/1 M) x (0.13 M) x (1 M)
k = 8.22 x 10⁻²M⁻² s⁻¹
Similarly, for trial 3:
[W] = 0.13 M
Rate = 2.95 x 10⁻¹ M/s
Again, using trial 1 as the reference trial, figure out the reaction's rate constant (k) in relation to U and S:
k = (Rate3/Rate1) x (1/[U]3) x (1/[S]3) x ([U]1) x ([S]1)
k = (2.95 x 10⁻¹ M/s) / (4.72 x 10⁻⁴ M/s) x (1/3.25 M) x (1/1 M) x (0.13 M) x (1 M)
k = 1.10 M⁻² s⁻¹
Therefore, the equation states the reaction rate in relation to U and S is k = 8.22 x 10⁻² M⁻² s⁻¹ and 1.10 M⁻² s⁻¹ for trials 2 and 3, respectively.
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If a balloon containing a mixture of gases where the partial pressure of oxygen is 12.8 psi, the partial pressure of hydrogen is 4.31 psi, and the partial pressure of nitrogen is 22.1 psi, then the actual pressure inside the balloon is…
Answers
Answer:
39.21 psi
Explanation:
According to Dalton's Law, the total pressure in the system is the sum of all the partial pressures, so all you need to do is add the partial pressures :)
What is the number of molecules of NO, which contains 16 gm of oxygen. 14
Answers
To find the number of molecules of NO that contains 16 g of oxygen, we need to first calculate the number of moles of oxygen in 16 g of oxygen. Using the atomic weight of oxygen (16 g/mol), we can calculate:
moles of O = mass of O / atomic weight of O = 16 g / 16 g/mol = 1 mol
Next, we need to determine the number of moles of NO that contains 1 mol of oxygen. From the molecular formula of NO, we can see that 1 mol of NO contains 1 mol of oxygen. Therefore, the number of moles of NO that contains 1 mol of oxygen is also 1 mol.
Finally, we can use Avogadro's number to convert the number of moles of NO to the number of molecules of NO. Avogadro's number is approximately 6.02 x 10^23 molecules/mol. Therefore, the number of molecules of NO that contains 16 g of oxygen is:
number of molecules of NO = number of moles of NO x Avogadro's number
number of molecules of NO = 1 mol x 6.02 x 10^23 molecules/mol
number of molecules of NO = 6.02 x 10^23 molecules
Therefore, there are approximately 6.02 x 10^23 molecules of NO that contain 16 g of oxygen.
aHow does the electronic configuration of a sodium cation differ from that of a sodium atom?
Answers
Answer:
Atomic number of sodium is 11
Electronic configuration of a sodium atom :
1s² 2s² 2p⁶ 3s¹Since sodium has one electron in its outermost shell, Therefore, sodium can easily donate it's one electron. As the result it becomes sodium cation with + 1 charge.
Electronic configuration of a sodium cation,[tex] \: \sf ({Na}^{+1}) [/tex]
1s² 2s² 2p⁶In case of sodium cation, it has fully filled electronic configuration.
Cations - Atoms that carry postive charge are called cations. Cations are formed when an atom loses its electron.
For example : [tex]\sf {Na}^{+} [/tex]
Anions - Atoms that carry negative charge are called anions. Anions are formed when an atom gains a electron.
For example : [tex]\sf {Cl}^{-} [/tex]
Chemistry what is the reaction Rate TABLE
Answers
The rate constant, k = 5.27 E-2 s⁻¹, determines the rate law for the reaction P → E + Z.
How to determine rate constant?The rate of the reaction P → E + Z can be expressed as:
Rate = - d[P]/dt = d[E]/dt = d[Z]/dt
where d[P], d[E], and d[Z] = changes in the concentrations of P, E, and Z, respectively, over a small time interval dt.
Use the experimental data to determine the rate constant and the order of the reaction.
Calculate the initial rate of the reaction in each trial by dividing the change in concentration of P by the time interval:
rate1 = (d[P]/dt)1 = (0.30 M - 0 M)/(20 s) = 0.015 M/s
rate2 = (d[P]/dt)2 = (0.60 M - 0.30 M)/(20 s) = 0.015 M/s
rate3 = (d[P]/dt)3 = (0.90 M - 0.60 M)/(20 s) = 0.015 M/s
The initial rates are the same in all three trials, which suggests that the reaction is first-order with respect to P.
Now using any of the three trials to determine the value of the rate constant k, trial 1:
Rate1 = k[P]1
k = Rate1/[P]1 = (1.58 E-2 M/s)/(0.30 M) = 5.27 E-2 s⁻¹
Therefore, the rate law for the reaction P → E + Z is:
Rate = k[P]
where k = 5.27 E-2 s⁻¹ is the rate constant.
Use the rate law to calculate the expected rates of the reaction at different concentrations of P. For example:
Rate2 = k[P]2 = (5.27 E-2 s⁻¹)(0.60 M) = 3.16 E-2 M/s
Rate3 = k[P]3 = (5.27 E-2 s⁻¹)(0.90 M) = 4.74 E-2 M/s
These expected rates are close to the experimental rates, which suggests that the rate law is a good approximation for the reaction under these conditions.
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Suppose a skimmer and a gull eat the same fish Over time the skimmer is more successful at catching the fish what would happen to each bird population
Pls help
Answers
If a skimmer and a gull eat the same fish and the skimmer is more successful at catching the fish over time, it is likely that the skimmer population would increase, while the gull population may decrease.
What is the skimmer's success in catching the fish?The skimmer's success in catching the fish would give it an advantage in obtaining the necessary nutrients for survival and reproduction. As a result, the skimmer population would likely grow over time as more individuals are able to survive and reproduce due to the abundance of food.
On the other hand, the gull population may decrease due to the competition with the skimmer for the same food source. If the skimmer population grows significantly, it may lead to a reduction in the availability of fish for the gulls to feed on. Over time, this could result in a decline in the gull population due to reduced food availability.
However, it is important to note that the impact on the bird populations may depend on various factors such as the size of the populations, availability of other food sources, and environmental factors. Therefore, the outcome of this scenario cannot be predicted with certainty and would require further analysis and investigation.
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In a reaction between vinegar and antacid tablets, the antacid is the limiting reagent. cm of gas. At constant pressure and temperature, three tablets produce 600 cm³ What volume will four tablets produce? 300 cm³ 600 cm³ 800 cm³ 3 1,200 cm³ 3
Answers
If in a reaction between vinegar and antacid tablets, the antacid is the limiting reagent. cm of gas. At constant pressure and temperature, three tablets produce 600 cm³ . The volume that four tablets will produce is: C. 800 cm³.
What volume will four tablets produce?Since the antacid is the limiting reagent, the amount of gas produced will be directly proportional to the number of tablets used.
We know that three tablets produced 600 cm³ of gas. Therefore, we can set up a proportion:
3 tablets produce 600 cm³ of gas
4 tablets produce x cm³ of gas
To solve for x, we can use cross-multiplication:
3 tablets × x cm³ of gas = 4 tablets × 600 cm³ of gas
3x = 2400
x = 800 cm³
Therefore the answer is C. 800 cm³.
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If the reaction A (aq) + B (aq) C(aq) has a Ka value equal to 4.26 x 10-6, what is the G value at 25 °C if the concentrations are as follows:
[A] = 1.50 M
[B] = 1.00 M
[C] = 5.00 x 10-5 M
Answers
The Gibbs free energy change for the given reaction at 25°C and the given concentrations is -25.5 kJ/mol
The Gibbs free energy change (∆G) of a reaction can be calculated using the equation:
∆G = -RT ln(K)
Where R is the gas constant (8.314 J/molK), T is the temperature in Kelvin, and K is the equilibrium constant.
The equilibrium constant (K) can be calculated from the acid dissociation constant (Ka) as:
K = [C] ÷ ([A] × [B])
Substituting the given values, we get:
K = (5.00 x 10⁻⁵) ÷ (1.50 x 1.00) = 3.33 x 10⁻⁵
Therefore,
∆G = - (8.314 J/molK) × (298 K) × ln(3.33 x 10⁻⁵)
= 25.5 kJ/mol
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. A ring with a mass of 25.5 g appears to be pure silver. Rather than test for density, you can confirm the ring's composition by determining its specific heat. Suppose the ring is heated to a temperature of 84.0°C and then immersed in a container of water until the ring's temperature is 25.0°C. If the ring gives up 667.5 J of energy to the water, what is its specific heat? Is the ring made of silver (C = 0.234 J/g °C), nickel (C = 0.444 J/g. °C), or palladium (C = 0.244 J/g °C) help me
Answers
The specific heat capacity of the ring, given that the ring gives up 667.5 J of energy to the water is 0.444 J/gºC. The ring is made of nickel.
How do i determine the specific heat capacity of the ring?The specific heat capacity of the ring can be obtain as illustrated below:
Heat absorbed by water (Q) = 667.5 JHeat released by ring (Q) = -667.5 JMass of ring (M) = 25.5 gInitial temperature of ring (T₁) = 84.0 °CFinal temperature (T₂) = 25.0 °CChange in temperature (ΔT) = 25.0 - 84.0 = -59 °CSpecific heat capacity of ring (C) = ?Q = MCΔT
-667.5 = 25.5 × C × -59
-667.5 = -1504.5 × C
Divide both sides by -1504.5
C = -667.5 / -1504.5
C = 0.444 J/gºC
Thus, the specific heat capacity of the ring is 0.444 J/gºC. Hence, the ring is made of nickel
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Calculate the volume of C2H2 that is collected over water at 23 ∘C by a reaction of 1.52 g of CaC2 if the total pressure of the gas is 751 torr. (The vapor pressure of water is 21.07 torr .)
Answers
The volume of C₂H₂ that is collected over water at 23 ∘C by the reaction of 1.52 g of CaC₂ is 0.61 L
How do i determine the volume of C₂H₂ collected?First, we shall determine the mole of CaC₂ that reacted. Details below:
Mass of CaC₂ = 1.52 g Molar mass of CaC₂ = 64 g/mol Mole of CaC₂ =?Mole = mass / molar mass
Mole of CaC₂ = 1.52 / 64
Mole of CaC₂ = 0.024 mole
Next, we shall determine the mole of C₂H₂. obtained. Details below:
CaC₂ + 2H₂O -> C₂H₂ + Ca(OH)₂
From the balanced equation above,
1 mole of CaC₂ reacted to produced 1 mole of C₂H₂
Therefore,
0.024 mole of CaC₂ will also react to produce 0.024 mole of C₂H₂
Finally, we shall determine the volume of C₂H₂ collected. This is shown below:
Temperature (T) = 23 °C = 23 + 273 = 296 KVapor pressure = 21.07 torrPressure of dry gas (P) = 751 - 21.07 = 729.93 torrGas constant (R) = 62.36 torr.L/mol KNumber of mole (n) = 0.024 moleVolume of gas (V) =?PV = nRT
729.93 × V = 0.024 × 62.36 × 296
Divide both sides by 729.93
V = (0.024 × 62.36 × 296) / 729.93
V = 0.61 L
Thus, the volume of the C₂H₂ gas collected is 0.61 L
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The iodine monobromide molecule, IBr, has a bond length of 249 pm and a dipole moment of 1.21 D. (a) Which atom of the molecule is expected to have a negative charge? (b) Calculate the effective charges on the I and Br atoms in IBr in units of the electronic charge, e.
Answers
a. Br will have the negative charge
b. The effective charges on the I and Br atoms are approximately +1.012e and -1.012e, respectively.
How to determine the negative chargea. To identify which atom within an IBr molecule will have a partial negative charge, we must consider each atom's electronegativity.
On the periodic table, iodine (I) has an electronegativity value of 2.66 while bromine (Br) boasts 2.96; since Br has higher electronegativity it will attract electrons more strongly and hence have an even stronger partial negative charge.
B. To calculate the effective charges on the I and Br atoms in IBr, we can use the dipole moment equation:
μ = Q * d
where μ is the dipole moment, Q is the effective charge, and d is the bond length.
We are given the dipole moment (μ) as 1.21 D, and the bond length (d) as 249 pm. However, we need to convert the units to the SI system before proceeding with the calculation.
[tex]1 D (Debye) = 3.336 * 10^-^3^0 cm,\\\\1 pm = 10^-^1^2 m.[/tex]
Now we can solve for the effective charge (Q):[tex]u = 1.21 D * (3.336 × 10^-^3^0 Cm/D)\\ \\= 4.03656 * 10^-^3^0 cm\\d = 249 pm * (10^-12 m/pm) = 2.49 * 10^-^1^0 m[/tex]
Q = μ / d
[tex]Q = (4.03656 * 10^-^3^0 cm) / (2.49 *10^-^1^0 m) \\\\\\=1.62151 * 10^-^2^0 C[/tex]
This effective charge represents the charge difference between the I and Br atoms. To express the charges in units of the elementary charge (e), we need to divide the effective charge by the elementary charge value (e = 1.602 × 10^-19 C):
Q_e =[tex]\frac{(1.62151 * 10^-^2^0 C)}{(1.602 * 10^-^1^9 C)} = 1.012[/tex]
The effective charges on the I and Br atoms are approximately +1.012e and -1.012e, respectively.
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What is the molarity of a solution that has 2.0 moles of solute in 3.0 L of solution?
Answers
The molarity of the solution that has 2.0 moles of solute in 3.0 L of solution is 0.67 mol/L
What is molarity?Molarity is described as a measure of the concentration of a chemical species, in particular of a solute in a solution, in terms of amount of substance per unit volume of solution.
Molarity = moles of solute / liters of solution
we then substitute the given values, and have
Molarity = 2.0 moles / 3.0 L
Molarity = 0.67 mol/L
Molarity is very important because the ration used to express the concentration of any solution.
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add a given instant what is the rate of appearance at this time
Answers
The rate of appearance of NOBr at the given time is 9.80 × [tex]10^-^4[/tex] M/s which is the second option as the question asks to determine the rate of appearance of NOBr, which is a product of the given chemical reaction, at the given time when the rate of disappearance of Br₂ is known.
For every 1 mole of Br₂ that disappears, 2 moles of NOBr appear. Thus, we can set up a proportion:
(2 mol NOBr / 1 mol Br₂) = (rate of appearance of NOBr / rate of disappearance of Br₂)
Substituting the given values,
(2 mol NOBr / 1 mol Br₂) = (rate of appearance of NOBr / 4.90 x [tex]10^-^4[/tex] M/s)
Solving for the rate of appearance of NOBr,
rate of appearance of NOBr = (2 mol NOBr / 1 mol Br₂) x (4.90 x [tex]10^-^4[/tex] M/s) = 9.80 × [tex]10^-^4[/tex]M/s
The rate of appearance of NOBr at the given time is 9.80 × [tex]10^-^4[/tex] M/s.
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A gas‑filled weather balloon has a volume of 56.0 L
at ground level, where the pressure is 761 mmHg
and the temperature is 23.1 ∘C.
After being released, the balloon rises to an altitude where the temperature is −6.97 ∘C
and the pressure is 0.0772 atm.
What is the weather balloon's volume at the higher altitude?
Answers
We can use the combined gas law to determine the volume of the balloon at a higher altitude. The combined gas law relates the pressure, volume, and temperature of a gas:
(P1 x V1) / T1 = (P2 x V2) / T2
where P1, V1, and T1 are the pressure, volume, and temperature of the gas at the initial state, and P2, V2, and T2 are the pressure, volume, and temperature of the gas at the final state.
We are given the initial pressure (P1 = 761 mmHg), volume (V1 = 56.0 L), and temperature (T1 = 23.1 °C = 296.25 K) of the gas, and the final pressure (P2 = 0.0772 atm), and temperature (T2 = -6.97 °C = 266.18 K) of the gas. We can solve for V2, the final volume of the gas:
(P1 x V1) / T1 = (P2 x V2) / T2
V2 = (P1 x V1 x T2) / (P2 x T1)
V2 = (761 mmHg x 56.0 L x 266.18 K) / (0.0772 atm x 296.25 K)
V2 = 2,040 L (rounded to three significant figures)
Therefore, the volume of the weather balloon at the higher altitude is approximately 2,040 L.
1. Which metal is the most reactive? How do you know this?
2. Rank the metals in order of increasing reactivity.
3. Give the chemical equations for each single replacement reaction that took place.
4. Was Fe^3+ reduced? Of so what metal(s) acted as reducing agents?
Answers
2. The metals can be ranked in order of increasing reactivity as follows: Gold (Au), Silver (Ag), Copper (Cu), Mercury (Hg), Lead (Pb), Tin (Sn), Iron (Fe), Zinc (Zn), Aluminum (Al), Magnesium (Mg), Sodium (Na), Potassium (K), Calcium (Ca).
3. The chemical equations for each single replacement reaction that took place are:
a. Zinc (Zn) + Copper (II) sulfate (CuSO4) → Zinc sulfate (ZnSO4) + Copper (Cu)
b. Iron (Fe) + Copper (II) sulfate (CuSO4) → Iron (II) sulfate (FeSO4) + Copper (Cu)
c. Aluminum (Al) + Copper (II) sulfate (CuSO4) → Aluminum sulfate (Al2(SO4)3) + Copper (Cu)
4. Yes, Fe^3+ was reduced to Fe^2+. The reducing agents were Zinc (Zn), Iron (Fe), and Aluminum (Al) which all have a higher reactivity than Fe.
Oxygen and oxygen-containing compounds are involved in many different reactions. Which of the following equations represent a balanced reaction involving 14 atoms of oxygen? Question 5 options: NH4Cl + KOH --> NH3 + H2O + KCl 2Na + 2H2O --> 2NaOH + H2 2C2H6 + 7O2 --> 4CO2 + 6H2O 4Fe + 3O2 --> 2Fe2O3
Answers
The equation that represents a balanced reaction involving 14 atoms of oxygen is:
2C2H6 + 7O2 --> 4CO2 + 6H2O
a. The relationship between variables in this equation is that 2 moles of C2H6 (ethane) react with 7 moles of O2 (oxygen) to produce 4 moles of CO2 (carbon dioxide) and 6 moles of H2O (water). This equation follows the law of conservation of mass, where the total number of atoms on both sides of the equation is the same, indicating a balanced reaction.
b. The graph is linear, as the coefficients of the reactants and products in the equation are whole numbers and form a consistent ratio. The coefficients of 2, 7, 4, and 6 represent the stoichiometry of the reaction, indicating a fixed relationship between the reactants and products.
c. An example of a situation where this balanced equation could be applicable is the combustion of ethane (C2H6) in the presence of excess oxygen (O2) to produce carbon dioxide (CO2) and water (H2O), which is a common reaction in the combustion of hydrocarbon fuels. The equation represents the balanced stoichiometry of this reaction, where 2 moles of ethane react with 7 moles of oxygen to produce 4 moles of carbon dioxide and 6 moles of water, involving a total of 14 atoms of oxygen in the reaction.
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A specific organic reaction is described by the energy diagram drawn below. Using this
energy diagram, identify which product will form first and which product will be the
major product if given enough time?
Answers
The product that is formed first is product B and the product that will be the major product if given enough time would also be product B.
In an energy diagram, the vertical axis represents the overall energy of the reactants, while the horizontal axis is the ‘reaction coordinate’, tracing from left to right the progress of the reaction from starting compounds to final products.
The activation energy of the reaction can be shown on a diagram as the energy between the reactants that the transition state.
The product with lesser activation energy is the product that is formed first and the major product is decided by the stability of the product which depends on the energy. Lesser is the energy of the product, greater is its stability.
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Potassium superoxide, KO2, reacts with carbon dioxide to form potassium carbonate and oxygen:
This reaction makes potassium superoxide useful in a self-contained breathing apparatus. How much O2 could be produced from 2.61 g of KO2 and 4.46 g of CO2?
Answers
First, we need to write out the balanced chemical equation for the reaction: 4 KO2 + 2 CO2 → 2 K2CO3 + 3 O2
From the equation, we can see that 4 moles of KO2 react with 2 moles of CO2 to produce 3 moles of O2. Therefore, we need to convert the given masses of KO2 and CO2 into moles:
moles of KO2 = 2.61 g / molar mass of KO2 = 2.61 g / 71.10 g/mol = 0.0367 mol
moles of CO2 = 4.46 g / molar mass of CO2 = 4.46 g / 44.01 g/mol = 0.1013 mol
Next, we need to determine the limiting reagent (the reactant that will be completely consumed in the reaction) by comparing the mole ratios of KO2 and CO2 in the balanced equation. The ratio of moles of KO2 to moles of CO2 is:
0.0367 mol KO2 / 4 mol KO2 per 2 mol CO2 = 0.0184 mol CO2
Since this ratio is less than the actual number of moles of CO2 we have (0.1013 mol), CO2 is in excess and KO2 is the limiting reagent.
Using the mole ratio from the balanced equation, we can calculate the number of moles of O2 produced:
moles of O2 = 3 mol O2 per 4 mol KO2 × 0.0367 mol KO2 = 0.0275 mol O2
Finally, we can convert the moles of O2 to grams:
mass of O2 = moles of O2 × molar mass of O2 = 0.0275 mol × 32.00 g/mol = 0.88 g
Therefore, 2.61 g of KO2 and 4.46 g of CO2 would produce 0.88 g of O2.
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Acetic acid has a molar mass of 60.05 g/mol. If 16.84 grams of acetic acid are present, how many moles of acetic acid does that correspond to?
Answers
Answer:
3.566 mol
Explanation:
Since 60.05 is grams divided by mol to cancel out the grams to get only mols it must be divided by 16.84 g
[tex]\frac{60.05 g}{mol} *\frac{1 }{16.84g} =3.566[/tex] mols acetic acid
in an experiment, 1 mol A, 2 mol B and 1 mol D were mixed and allowed to come to equilibrium at 25C. The resulting mixture was found to contain 0.9 mol of C at a total pressure of 1.00 bar. Find the mole fractions of each species at equilibrium
Answers
The mole fractions of each species at equilibrium are 0.25 for A and D, 0.5 for B, and 0.225 for C.
we can use the principles of chemical equilibrium and the mole fraction formula.
First, we need to write the balanced chemical equation for the reaction involving A, B, C, and D. Let's assume that the reaction is:
A + 2B <=> C + D
where A, B, C, and D are the chemical species, and the coefficients indicate their stoichiometric ratios.
Next, we need to write the expression for the equilibrium constant, Kc, for this reaction:
Kc = [C][D] / [A][B]²
where [X] denotes the molar concentration of species X at equilibrium.
Since we know the initial moles of A, B, and D, we can calculate their total moles in the mixture:
Total moles = 1 mol A + 2 mol B + 1 mol D = 4 mol
We also know that the final mixture contains 0.9 mol of C. Therefore, the molar concentration of C at equilibrium is:
[C] = 0.9 mol / 4 L = 0.225 M
Since we have only one unknown, we can use the equilibrium constant expression to calculate the molar concentration of D:
Kc = [C][D] / [A][B]²
0.9 = (0.225)(D) / (1)(2²)
D = 1.8
Therefore, the molar concentration of D at equilibrium is 1.8 M.
Using the law of conservation of mass, we can also calculate the molar concentration of A and B at equilibrium:
[A] = 1 mol / 4 L = 0.25 M
[B] = 2 mol / 4 L = 0.5 M
Mole fraction of X = moles of X / total moles
Mole fraction of A = 1 mol / 4 mol = 0.25
Mole fraction of B = 2 mol / 4 mol = 0.5
Mole fraction of C = 0.9 mol / 4 mol = 0.225
Mole fraction of D = 1 mol / 4 mol = 0.25
Therefore, the mole fractions of each species at equilibrium are 0.25 for A and D, 0.5 for B, and 0.225 for C.
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how many moles of a salute are needed to prepare 300 mL of 0.8 mm and NACL solution
Answers
Answer: 0.24 Moles. WARNING: "mm" in the original question seems to be a typo and was assumed to be "M" (moles per liter) for concentration.
Explanation:
To calculate the number of moles of solute needed to prepare a solution, we can use the formula:
moles of solute = concentration (in moles per liter) * volume of solution (in liters)
Given:
Concentration of NaCl solution = 0.8 M (moles per liter)
Volume of solution to be prepared = 300 mL = 300/1000 L (converted to liters)
Plugging in the given values:
Concentration = 0.8 M
Volume of solution = 300/1000 L
moles of NaCl = 0.8 M * 300/1000 L
Calculating:
moles of NaCl = 0.24 moles
So, 0.24 moles of NaCl are needed to prepare 300 mL of a 0.8 M NaCl solution.
What is the molar equilibrium concentration of Cu2+ in a solution where 1.00 L of 0.800 M [Cu(NH3)4]2+ has 0.100 mol NH3 added to it, without any change in volume. The equation is Cu2+(aq) + 4 NH3(aq) [Cu(NH3)4]2+(aq) , where Kf = 2.1 x 1013
Answers
The molar equilibrium concentration of Cu² ⁺ is [tex]3.3 x 10^-26[/tex]M when 0.100 mol NH₃ is added to 1.00 L of 0.800 M with Kf = [tex]2.1 x 10^13[/tex].
The issue includes computing the molar balance convergence of Cu² ⁺ in an answer that has 0.100 mol NH3 added to 1.00 L of 0.800 M [Cu(NH₃)₄]²⁺ . To tackle the issue, we can involve the balance consistent articulation for the arrangement of[Cu(NH₃)₄]²⁺:
Kf =[Cu(NH₃)₄]²⁺/(Cu² ⁺)(NH₃)₄
Since the volume of the arrangement doesn't change when NH₃ is added, the underlying grouping of [Cu(NH₃)₄]²⁺ is as yet 0.800 M, while the underlying convergences of Cu² ⁺ and NH₃ are zero. Let x be the molar centralization of Cu² ⁺ at balance, and (0.100 - 4x) be the molar grouping of NH₃ at harmony. Then, we can compose the harmony consistent articulation as:
[tex]2.1 x 10^13 = (x)(0.800 - x)^4/(0.100 - 4x)^4[/tex]
Settling for x gives the molar balance convergence of Cu² ⁺ as 3.3 x [tex]10^-26[/tex] M. This tiny focus demonstrates that the majority of the copper in the arrangement is as [Cu(NH₃)₄]²⁺ and that the expansion of NH₃ has moved the harmony towards the development of this complex.
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What happens to a buffered solution when a small amount of base is added?
O The solution quickly becomes neutral.
O The solutions resists changes in pH.
O The solution slowly becomes acidic.
O The solution quickly becomes basic.
Answers
Answer:
solution resists changes in pH
Explanation:
the inherent property of buffers is to resist change to ph even when acids and bases are added. when the base is added, it is quickly neutralized by the conjugate acid, so the ph won't change.
Answer:
B: The solutions resists changes in pH.
Explanation:
Buffer reactions maintain stable pH of solutions.
The 500 cm³ of a pas enclosed in a container under a pressure of 580 mm of Hg. If the volume is reduced to 300 cm³ what will be the pressure then?
Answers
Answer:
The answer is 966.67 mm of Hg.
Explanation:
To solve this problem, we can use Boyle's Law which states that the pressure of a gas is inversely proportional to its volume when the temperature and the amount of gas are kept constant. The formula for Boyle's Law is:
P1V1 = P2V2
where P1 is the initial pressure, V1 is the initial volume, P2 is the final pressure, and V2 is the final volume.
Using the given values:
P1 = 580 mmHg
V1 = 500 cm³
V2 = 300 cm³
We can solve for P2:
P1V1 = P2V2
580 mmHg x 500 cm³ = P2 x 300 cm³
290,000 mmHg·cm³ = P2 x 300 cm³
P2 = 290,000 mmHg·cm³ / 300 cm³
P2 = 966.67 mmHg (rounded to the nearest hundredth)
Therefore, the pressure when the volume is reduced to 300 cm³ is approximately 966.67 mmHg.
The compounds labeled benzophenone-3 (C14H12O3) and benzophenone-5 (C14H11NaO6S) are found in certain sunscreens. Would you expect a sunscreen made with benzophenone-3 or benzophenone-5 be more waterproof? Explain your choice.
Answers
A sunscreen made with Benzophenone-5 ([tex]C_1_4H_1_1NaO_6S[/tex]) would be expected to be more waterproof than benzophenone-3 ([tex]C_1_4H_1_2O_3[/tex]).
This is due to the presence of a sodium salt group (Na) and a sulfonic acid group ([tex]SO_3H[/tex] ) in benzophenone-5, which makes it more polar than benzophenone-3. Polar molecules interact more strongly with water molecules and are less likely to dissolve in nonpolar solvents such as oils.
Because sunscreen is designed to be water-resistant, the more polar benzophenone-5 should have stronger interactions with water and give more water resistance than benzophenone-3.
Moreover, the sulfonic acid group in benzophenone-5 may allow it to make stronger hydrogen bonds with water, increasing its water resistance even further.
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How many grams of NiNO can be produced if 35.1 g of ammonium nitrate and 552 g of sodium phosphate react?
Answers
Answer:its b
Explanation:
jusut bcss
your answer to the following question on the information below and you knowledge of chemistry.
A 100. -gram sample of liquid water is heated from 30.0°C to 80.0°C. Enough KCIO:(s) is dissolved in the sample of water at 80.0°C to form a saturated solution.
Based on Table H, determine the vapor pressure of the water sample at its final temperature.
Answers
Explanation:
Table H lists vapor pressure data for pure water at various temperatures. We can use this data to estimate the vapor pressure of the water in the given system at its final temperature of 80.0°C.
First, we need to calculate the heat absorbed by the water sample during the heating process. We can use the specific heat capacity of water to do this:
q = m * c * ΔT
where q is the heat absorbed, m is the mass of water (100 g), c is the specific heat capacity of water, and ΔT is the temperature change (80°C - 30°C = 50°C).
Plugging in the values, we get:
q = 100 g * 4.18 J/(g*C) * 50 C
q = 20900 J
This tells us that 20,900 joules of energy were absorbed by the water sample during heating.
Next, we need to consider the saturated solution of KCIO3 in the water sample. At 80.0°C, the water is already close to boiling, so it is likely that the vapor pressure of the water in the system is close to the vapor pressure of pure water at this temperature. From Table H, we can see that the vapor pressure of pure water is approximately 356 mmHg at 80.0°C.
Therefore, the vapor pressure of the water in the given system at its final temperature of 80.0°C is approximately 356 mmHg.
Given the following reaction:
2C₂H₂(g) + 5O₂(g) → 4CO₂(g) + 2H₂O(g) ∆H = -2511.6 kJ
What is the energy change when 7.76 g of C₂H₂ react with excess O₂?
Answers
Chemical reactions nearly always include a change in energy between the products and reactants.
Thus, Energy is released when chemical bonds are created and is released when chemical bonds are destroyed.
The overall energy of a system, however, must remain constant according to the Law of Conservation of Energy, and chemical reactions frequently absorb or release energy in the form of heat, light, or both.
The difference in the amounts of chemical energy that are stored in the products and reactants accounts for the energy change in a chemical reaction. Enthalpy refers to the system's heat content or stored chemical energy.
Thus, Chemical reactions nearly always include a change in energy between the products and reactants.
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What mass of oxygen would be released by the thermal decomposition of 918.7 grams of Mercury (II) Oxide?
HgO --> Hg + O2
Answers
Answer:
Explanation:
[tex]\frac{918.7 g}{1} *\frac{1}{216.59m } = 4.241 mol[/tex] To start off the mol of HgO must be found.
After that the molar ratio between HgO and O must be found but in this case its 1:1
[tex]4.241 mol HgO*\frac{1 molO}{1molHgO} = 4.241 mol O[/tex] the mols of HgO is put on the bottom to cancel out with the other one leaving just mols of oxygen. Finally to find g of oxygen it must be multiplied by its molar mass.
[tex]\frac{4.241 molO}{1} * \frac{15.999 g}{mol} = 67.85 g[/tex] Oxygen
