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photoshop1234 [79]
3 years ago
13

The catalyst used in the industrial production of ammonia: Select the correct answer below: A. enables equilibrium to be reached

more quickly B. increases the equilibrium constant C. shifts the equilibrium of the reaction to favor the products D. none of the above
Chemistry
1 answer:
max2010maxim [7]3 years ago
8 0

Answer: A. enables equilibrium to be reached more quickly..

Explanation:

The catalyst can be define as an important ingredient may be a protein in the form of enzyme that is used to speed up the rate of the chemical reaction. The catalyst cause a transformational change in the substrate to produce product. The catalyst do not get consumed in the reaction. In the availability of the catalyst, the forward and reverse rates of the reactions will speed up this will allow the equilibrium to reach faster.

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Kevin has 5 fish in his fish tank jasmine has 4 times as many fish as Kevin.How many fish does jasmine have?
madreJ [45]
Jasmine have 20 fish because4 times 5 equals 20
4 0
3 years ago
When 5.0 mol Al react with 13 mol HCl, what is the limiting reactant, and how many moles of AlCl3 can be formed? 2Al + 6HCl → 2A
bekas [8.4K]
HCI is the limiting reactant, and 4.3 mol AICI3 can be formed.
6 0
3 years ago
What is the total number of electron pairs shared between the carbon atom and one of the oxygen atoms in a carbon dioxide molecu
TEA [102]
The answer is 2 pairs. The carbon atom has 4 electrons and oxygen atom has 6 electrons. The purpose of shared electron is to get 8 electrons at the outer layer. So the carbon atom needs to share 4 electrons which is 2 pairs with oxygen to be stable.
7 0
3 years ago
Cu(s) + 4 HNO3 (aq) --> Cu(NO3)2 (aq) + 2NO2 (g) + 2H2O(l)
GREYUIT [131]

Answer:

The percent by mass of copper in the mixture was 32%

Explanation:

The ammount of HNO₃ used is:

mol HNO₃ = volume * concentration

mol HNO₃ = 0.015 l * 15.8 mol/l

mol HNO₃ = 0.237 mol

According to the reaction, 4 mol HNO₃ will react with 1 mol Cu and produce 1 mol Cu²⁺. Since we have 0.237 mol HNO₃, the amount of Cu that could react would be (0.237 mol HNO₃ * 1 mol Cu / 4 mol HNO₃) 0.06 mol. This reaction would produce 0.060 mol Cu²⁺, however, only 0.010 mol Cu²⁺ were obtained, indicating that only 0.010 mol Cu were present in the mixture. This means that the acid was in excess, so we can assume that all copper present in the mixture has reacted.

Since 0.010 mol of Cu²⁺ were produced, the amount of Cu was 0.01 mol.

1 mol of Cu has a mass of 63.5 g, then 0.01 mol has a mass of:

0.01 mol Cu * 63.5 g / 1 mol = 0.635 g.

Since this amount was present in 2.00 g mixture, the amount of copper in 100 g of the mixture will be:

100 g(mixture) * 0.635 g Cu / 2 g(mixture) = 32 g

Then, the percent by mass of Cu (which is the mass of Cu in 100 g mixture) is 32%

6 0
3 years ago
Read 2 more answers
How many grams of propane are in 20 pounds of propane? Use the conversion 1 lb = 454 g. (Express your answers for the next three
Dima020 [189]

Answer:

a. =9.1x10^3gC_3H_8

b. 2.1x10^2molC_3H_8

c. Q=-4.6x10^5kJ

Explanation:

Hello,

a. By applying the given information, one obtains:

20lbC_3H_8*\frac{454gC_3H_8}{1lbC_3H_8} =9.1x10^3gC_3H_8

b. By knowing that the propane has a molecular mass of 44g/mol, one obtains:

20lbC_3H_8*\frac{454gC_3H_8}{1lbC_3H_8}*\frac{1molC_3H_8}{44gC_3H_8} =2.1x10^2molC_3H_8

c. Here, the propane's combustion chemical reaction is stated:

C_3H_8+5O_2-->3CO_2+4H_2O

This enthalpy of reaction is computed via:

ΔrH=(3*-393.5kJ/mol)+(4*241kJ/mol)-(-104.7kJ/mol)=-2043kJ/mol

Finally, since it is done for 20 lb of propane (2.1x10^2molC_3H_8), the obtainable energy is:

Q=-2043kJ/mol*2.1x10^2molC_3H_8\\Q=-4.6x10^5kJ

Best regards.

4 0
3 years ago
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