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hram777 [196]
3 years ago
11

Which increases the rate of a chemical reaction? A. a catalyst B. a higher volume C. a lower pressure D. a lower temperature

Chemistry
2 answers:
Tanya [424]3 years ago
6 0
A. A catalyst because it increases the time that it takes to happen
taurus [48]3 years ago
3 0
A catalyst will increase the rate of a chemical reaction by changing the reaction mechanism so that the reaction has a lower activation energy.  

a higher volume will decrease the concentrations which will intern decrease the number of collisions which will slow down the rate of the reaction since collisions between reactants are what make reactions happen.  A lower temperature causes the molecules to slow down thereby lowering the collision rate which will lower the reaction rate.  A lower pressure will lower the number of collisions which will intern result in a lower reaction rate (this can be thought of as a lower concentration since lower pressures involve less gas per unit volume)
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3 years ago
3. Mr. Hill has 27 students in his class
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3 years ago
Suppose 12.0 g of carbon (c) reacts with 70.0 g of sulfur (s) to give 76.0 g of the compound carbon disulfide 1 c s 2 2 . In the
Nitella [24]

Answer is: mass of unused sulfur is 5.87 grams.

Balanced chemical reaction: C + 2S → CS₂.

m(C) = 12.0 g; mass of carbon.

m(S) = 70.0 g; mass of sulfur.

n(C) = m(C) ÷ M(C).

n(C) = 12 g ÷ 12 g/mol.

n(C) = 1 mol; amount of substance.

n(S) = m(S) ÷ M(S).

n(S) = 70 g ÷ 32.065 g/mol.

n(S) = 2.183 mol.

From chemical reaction: n(C) : n₁(S) = 1 : 2.

n₁(S) = 1 mol · 2 = 2 mol.

Δn(S) = n(S) - n₁(S).

Δn(S) = 2.183 mol - 2 mol.

Δn(S) = 0.183 mol; amount of unused sulfur.

Δm(S) = 0.183 mol · 32.065 g/mol.

Δm(S) = 5.87 g.

4 0
3 years ago
Imagine that you have a 5.00 L gas tank and a 3.50 L gas tank. You need to fill one tank with oxygen and the other with acetylen
Lorico [155]

Answer:

77.14 atm of pressure should be of an acetylene in the tank.

Explanation:

2C_2H_2+5O_2\rightarrow 4CO_2+2H_2O

According to reaction, 2 moles of acetylene reacts with 5 moles of oxygen.

Moles of oxygen=n_1

Moles of acetylene =n_2

\frac{n_1}{n_2}=\frac{5 mol}{2 mol}=\frac{5}{2}

Volume of large tank with oxygen gas, V_1 = 5.00 L

Pressure of the oxygen gas inside the tank = P_1=135 atm

RT=\frac{P_1V_1}{n_1} ..[1]

Volume of small tank with acetylene gas ,V_2= 3.50 L

Pressure of the acetylene gas inside the tank = P_2=?

RT=\frac{P_2V_1}{n_2} ..[2]

Considering both the gases having same temperature T, [1]=[2]

\frac{P_1V_1}{n_1}=\frac{P_2V_2}{n_2}

P_2=\frac{P_1V_1\times n_2}{V_2\times n_1}

=\frac{135 atm\times 5.00 L\times 2}{3.50 L\times 5}=77.14 atm

77.14 atm of pressure should be of an acetylene in the tank.

8 0
3 years ago
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