Answer:
Explanation:
We have the reactions:
A: 
B: 
Our <u>target reaction</u> is:
We have 
as a reactive in the target reaction and is present in A reaction but in the products side. So we have to<u> flip reaction A</u>.
A: 
Then if we add reactions A and B we can obtain the target reaction, so:
A:
B: 
For the <u>final Kc value</u>, we have to keep in mind that when we have to <u>add chemical reactions</u> the total Kc value would be the <u>multiplication</u> of the Kc values in the previous reactions.
First, you want to extract the negative from -log(x).
So now you have log(x) = -2
Now you have to use the property loga(x)=b is the same as x=a^2
So now, it is x = 10^-2. Remember that when there is just a log, it is implied that it is ‘a’ is 10.
Then you evaluate the negative square to 1/10^2
Answer is 1/100
The grams of glucose are needed to prepare 400g of a 2.00%(m/m) glucose solution g is calculated as below
=% m/m =mass of the solute/mass of the solution x100
let mass of solute be represented by y
mass of solution = 400 g
% (m/m) = 2% = 2/100
grams of glucose is therefore =2/100 = y/400
by cross multiplication
100y = 800
divide both side by 100
y= 8.0 grams
