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baherus [9]
3 years ago
5

Vaping is... A. ...Holding an e-cigarette or vape device to look cool. B. ...The act of inhaling and exhaling tobacco from a cig

arette C. ...The act of inhaling and exhaling water vapor only. D. ...Using an e-cigarette or vape device to inhale the vapor.
Chemistry
1 answer:
Lelechka [254]3 years ago
5 0

Answer:

D

Explanation:

Process of elimination. There is more than just water vapor in a vape so it can't be D, it isn't a cigarette so it can't be B, and A is just clearly not the answer haha.

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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

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