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dexar [7]
3 years ago
13

Adding a catalyst to a system at equilibrium lowers the activation energy required by a system, which system, which shifts the e

quilibrium position toward the products.
True or false?
Chemistry
2 answers:
GarryVolchara [31]3 years ago
8 0

Answer: False

Explanation: Took the test

Len [333]3 years ago
5 0

False is the correct answer.

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The reaction A + BC --&gt; B+ AC is a<br> reaction.
jekas [21]

Explanation:

its answer is displacement reaction

4 0
3 years ago
At a certain concentration of H2 and I2, the initial rate of reaction is 4.0 x 104 M / s. What would the initial rate of the rea
andreev551 [17]

The question is incomplete, here is the complete question:

The rate of certain reaction is given by the following rate law:

rate=k[H_2]^2[I_2]^2

At a certain concentration of H_2 and I_2, the initial rate of reaction is 4.0 × 10⁴ M/s. What would the initial rate of the reaction be if the concentration of

Answer : The initial rate of the reaction will be, 1.0\times 10^4M/s  

Explanation :

Rate law expression for the reaction:

rate=k[H_2]^2[I_2]^2

As we are given that:

Initial rate = 4.0 × 10⁴ M/s

Expression for rate law for first observation:

4.0\times 10^4=k[H_2]^2[I_2]^2 ....(1)

Expression for rate law for second observation:

R=k(\frac{[H_2]}{2})^2[I_2]^2 ....(2)

Dividing 2 by 1, we get:

\frac{R}{4.0\times 10^4}=\frac{k(\frac{[H_2]}{2})^2[I_2]^2}{k[H_2]^2[I_2]^2}

\frac{R}{4.0\times 10^4}=\frac{1}{4}

R=1.0\times 10^4M/s

Therefore, the initial rate of the reaction will be, 1.0\times 10^4M/s

4 0
3 years ago
What do you know for sure after one test? Two tests? Three tests?
Zanzabum
That you have to do better
8 0
3 years ago
PLEASE HELP ME ANSWER THE TWO MULTIPLE CHOICE QUESTIONS!!!
Nikitich [7]

Answer:

4.D. 2KJ

5.B. close to 0 degree celsius

Explanation:

heat absorbed= heat given out

100g * 4.2 * (x-0)= 400g * 4.2* ( 20-x) where x is the new temp reached

then x= 20

i.e. the temperature of ice increases to 5 degree celsius and the temperature of water decreases to 5 degree celsius .

hope it helps

7 0
3 years ago
Ammonia, NH3 is a common base with Kb of 1.8 X 10-5. For a solution of 0.150 M NH3:
Vesnalui [34]

The concentrations : 0.15 M

pH=11.21

<h3>Further explanation</h3>

The ionization of ammonia in water :

NH₃+H₂O⇒NH₄OH

NH₃+H₂O⇒NH₄⁺ + OH⁻

The concentrations of all species present in the solution = 0.15 M

Kb=1.8 x 10⁻⁵

M=0.15

\tt [OH^-]=\sqrt{Kb.M}\\\\(OH^-]=\sqrt{1.8\times 10^{-5}\times 0.15}\\\\(OH^-]=\sqrt{2.7\times 10^{-6}}=1.64\times 10^{-3}

\tt pOH=-log[OH^-]\\\\pOH=3-log~1.64=2.79\\\\pH=14-2.79=11.21

8 0
3 years ago
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