Answer:
The molarity of the final solution is 1.7 M
Explanation:
The parameters given are;
First solution = 400 ml of 1.25 M
Second solution = 300 ml of 2.30 M
Therefore, we have;
First solution contains 400/1000 * 1.25 moles = 0.5 moles of the substance
Second solution contains 300/1000 * 2.30 moles = 0.69 moles of the
Hence the sum of the two solutions contains 0.5 + 0.69 = 1.19 moles of the substance
The volume of the sum of the two solutions = 400 ml + 300 ml = 700 ml
Hence we have the concentration of the final solution presented as follows;
700 ml contains 1.19 moles of the substance
Therefore;
1000 ml will contain 1000/700 * 1.19 = 1.7 moles
The molarity of the final solution = The number of moles per 1000 ml = 1.7 M.
The percent yield is given by the actual and the theoretical yield. The percent yield of the 2.73 gm compound is 80.3%. Thus, option b is correct.
<h3>What is the percent yield?</h3>
The percent yield is the ratio of the actual yield to the theoretical yield. It is calculated as:
% Yield = Actual yield ÷ Theoretical yield × 100%
= 2.73 ÷ 3.84 × 100%
= 71.09 %
As the closest number to 71.09 is 80.3% the percent yield will be 80.3%.
There is no limiting reagent in the reaction between the copper iodide and mercury iodide as the moles of the reactants and the products are in equal quantity.
Hence, option b. there is no limiting reagent in this reaction.
Learn more about percent yield here:
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Answer:
The equilibrium concentration of CH₃OH is 0.28 M
Explanation:
For the reaction: CO (g) + 2H₂(g) ↔ CH₃OH(g)
The equilibrium constant (Keq) is given for the following expresion:
Keq= 
=14.5
Where (CH3OH), (CO) and (H2) are the molar concentrations of each product or reactant.
We have:
(CH3OH)= ?
(CO)= 0.15 M
(H2)= 0.36 M
So, we only have to replace the concentrations in the equilibrium constant expression to obtain the missing concentration we need:
14.5= 
14.5 x (0.15 M) x 
= (CH₃OH)
0.2818 M = (CH₃OH)
