Molar mass Cu(OH)₂ = <span>97.561 g/mol
1 mol --------- 97.561 g
? mol ---------- 68 g
moles = 68 * 1 / 97.561
moles = 68 / 97.561
= 0.6969 moles
hope this helps!</span>
<span>If a solution conducts electricity, it it positive evidence that solute dissolved in solvent is electrolyte.</span>
Let's begin with the basic values that will be used in the solution.
The formula of propane is C3H8. It is an alkane, a hydrocarbon with the general formula of CnH2n+2. Notice that hydrocarbons have only Carbon and Hydrogen atoms. Its molar mass (M) is 44 g.
Molar Mass Calculation is done as like that
C=12 g/mol, H=1 g/mol. 1 mole propane has 3 moles Carbon atoms and 8 mole Hydrogen atoms. M(C3H8)= 3*12+ 8*1= 44 g
Combustion reaction of hydrocarbons gives carbon dioxide and water by releasing energy. That energy is called as enthalpy of combustion (ΔHc°).
ΔHc° of propane equals -2202.0 kj/mol. Burning of 1 mole C3H8 releases 2202 kj energy. Minus sign only indicates that the energy is given out ( an exothermic reaction ).
Let's write the combustion reaction.
C3H8 + O2 ---> CO2 + H20 (unbalanced) ΔHc° = -2202 kj/mol
Now, we calculate mole of 20 kg propane. Convert kilogram into gram since we use molar mass is defined in grams.
mole=mass/molar mass ; n=m/M ; n= 20000 g /44 (g/mol)=454 mole
1 mole propane releases 2202 kj energy.
454 mole propane release 2202 kj *454= 1000909 kj
The answer is 1000909 kj.
the complete question in attached figure
Let
x------------------- > actual yield
y------------------- > theoretical yield
z------------------- > percent yield
we have that
z=x/y
we know
x=47 g
y=56 g
therefore
z=47/56=0.839 ---------------- > 83.9%
the answer is the option C 83.9%
Answer:
i. Keq=4157.99.
ii. More hydrogen sulfide will be produced.
Explanation:
Hello,
i. In this case, for the concentrations at equilibrium on the given chemical reaction, the equilibrium constant results:
![Keq=\frac{[H_2S]^2}{[H_2]^2[S_2]} =\frac{(0.97M)^2}{(0.051M)^2(0.087)} =4157.99](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7B%5BH_2S%5D%5E2%7D%7B%5BH_2%5D%5E2%5BS_2%5D%7D%20%3D%5Cfrac%7B%280.97M%29%5E2%7D%7B%280.051M%29%5E2%280.087%29%7D%20%3D4157.99)
ii. Now, by means of the Le Chatelier's principle, the addition of a reactant shifts the reaction towards products, it means that more hydrogen sulfide will be produced in order to reach equilibrium.
Best regards.