Answer:
% composition O = 19.9%
% composition Cu = 80.1%
Explanation:
Given data:
Total mass of compound = 3.12 g
Mass of copper = 2.50 g
Mass of oxygen = 3.12 - 2.50 = 0.62 g
% composition = ?
Solution:
Formula:
<em>% composition = ( mass of element/ total mass)×100</em>
% composition Cu = (2.50 g / 3.12 g)×100
% composition Cu = 0.80 ×100
% composition Cu = 80.1%
For oxygen:
<em>% composition = ( mass of element/ total mass)×100</em>
% composition O = (0.62 g / 3.12 g)×100
% composition O = 0.199 ×100
% composition O = 19.9%
Explanation:
I hope this helps Chemistry is so hard and I hate it
The molarity of a solution that contains 35.00 g of CuSO4 dissolved in 250.0 mL of water is 0.88M.
<h3>How to calculate molarity?</h3>
The molarity of a solution can be calculated using the following formula:
Molarity = no of moles/volume
According to this question, a solution consists of 35.00 g of CuSO4 dissolved in 250.0 mL of water.
no.of moles of CuSO4 = 35g ÷ 159.6g/mol
no. of moles of CuSO4 = 0.22 moles
Therefore; molarity of CuSO4 solution is calculated as follows:
M = 0.22 ÷ 0.25
M = 0.88M
Therefore, the molarity of a solution that contains 35.00 g of CuSO4 dissolved in 250.0 mL of water is 0.88M.
Learn more about molarity at: brainly.com/question/12127540
The value of equilibrium constant is equal to the quotient of the products raised to its stoichiometric coefficient over the reaction's reactants raised to its respective stoichiometric coeff. The equation is Kc=[SO2][Cl2]/[SO2Cl2]= [1.3*10^-2][1.3*10^-2]/[2.2*10^-2-<span>1.3*10^-2]=0.0188. The final answer is Kc=0.0188.</span>
