Wednesday
April 1, 2015

Homework Help: Chemistry

Posted by Eliz on Monday, February 23, 2009 at 9:54pm.

Question: the rate constant for the second order decomposition of NOBr(g) is .800 1 divided by MS at 10 degree's C. Suppose a 2.50 L beaker contains NOBr at a concentration at .0680 moles per liter. How much time does it take for 75.0 percent of the sample to decompose.

I know that the equation is 1 divided by concentration of A=kt+1 divided by concentration of A(initially). I'm not sure how you get 1 divided by concentration of A. I know that k=.800 and that the initial concentration is 1 divided by(.0680) and that we are solving for t(time). How do you get the 1 divided by concentration of A with the information from the problem?

Answer this Question

First Name:
School Subject:
Answer:

Related Questions

Chemistry - Given the elementary reaction: 2 NOBr==>2 NO + Br2 k = 0.80 Which...
Chemistry - Nitrosyl bromide, NOBr, readily dissociated according to the ...
chemistry 1310 - The decomposition of nitrosyl bromide is followed by measuring ...
Physical Chemistry - At 25.0 degree celcius, the equilibrium 2 NOBr (gas) -->...
AP Chemistry - Consider the following reaction: 2NOBr(g) 2NO(g) + Br2(g) If 0....
Chemistry - 1. Nitric oxide reacts with bromine gas at elevated temperatures ...
chemistry - Nitric oxide reacts with bromine gas at elevated temperatures ...
Chemistry - Determine Kc for the following reaction: 1/2N2(g) + 1/2O2(g)+ 1/2Br(...
Chemistry - A 100 L reaction container is charged with 0.782 mol of NOBr, which ...
chemistry - nitrosyl bromide, NOBr is formed from NO and Br2-. Experiments show ...

Members