Answer:
0.209 mol/L
Explanation:
Given data
- Mass of copper(lI) sulfate (solute): 11.7 g
- Volume of solution: 350 mL = 0.350 L
The molar mass of copper(Il) sulfate is 159.61 g/mol. The moles corresponding to 11.7 grams are:
11.7 g × (1 mol/159.61 g) = 0.0733 mol
The molarity of copper(Il) sulfate is:
M = moles of solute / liters of solution
M = 0.0733 mol / 0.350 L
M = 0.209 mol/L
Answer:
-1,103.39KJ/mol
Explanation:
We use the subtract the standard enthalphies of formation of the reactants from that of the products. It must be taken into consideration that the enthalpy of formation of elements and their molecules alone are not taken into consideration. Hence, what we would be considering are the standard enthalpies of formation of H2S, H2O and SO2.
In places where we have more than one mole, we multiply by the number of moles as seen in the balanced chemical equations.
The standard enthalpies of the molecules above are as follows:
H2S = -20.63KJ/mol
H2O = -285.8KJ/mol
SO2 = -296.84KJ/mol
O2 = 0KJ/mol
ΔrH⦵ = [2ΔfH⦵(H2O) + 2 ΔfH⦵(SO2)] − [ΔfH⦵(H2S) + 3
ΔfH⦵(O2)]
ΔrH⦵ =[(2 × -285.8) + (2 × -296.84)]
-[ 3 × -20.63)]
= (-571.6 - 593.68 + 61.89) = -1,103.39KJ/mol
Stoichiomety:
1 moles of C + 1 mol of O2 = 1 mol of CO2
multiply each # of moles times the atomic molar mass of the compund to find the relation is weights
Atomic or molar weights:
C: 12 g/mol
O2: 2 * 16 g/mol = 32 g/mol
CO2 = 12 g/mol + 2* 16 g/mol = 44 g/mol
Stoichiometry:
12 g of C react with 32 g of O2 to produce 44 g of CO2
Then 18 g of C will react with: 18 * 32/ 12 g of Oxygen = 48 g of Oxygen
And the result will be 12 g of C + 48 g of O2 = 60 g of CO2.
You cannot obtain 72 g of CO2 from 18 g of C.
May be they just pretended that you use the law of consrvation of mass and say that you need 72 g - 18g = 54 g. But it violates the proportion of C and O2 in the CO2 and is not possible.
Answer:
Percent yield of SiC is 77.0%.
Explanation:
Balanced reaction: 
Molar mass of SiC = 40.11 g/mol
Molar mass of 
= 60.08 g/mol
So, 100.0 kg of = 
moles of = 1664 moles of
According to balanced equation, 1 mol of produces 1 mol of SiC
Therefore, 1664 moles of produce 1664 moles of SiC
Mass of 1664 moles of SiC = 
= 66743g = 66.74 kg (4 sig. fig.)
Percent yield of SiC = [(actual yield of SiC)/(theoretical yield of SiC)]
%
= 
%
= 77.0%
2 electrons. it has electron configuration of 2.8.8.2 hence the last 2 will be lost for it to gain stable.
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