Answer:
398 mL
Explanation:
Using the equation for molarity,
C₁V₁ = C₂V₂ where C₁ = concentration before adding water = 8.61 mol/L and V₁ = volume before adding water, C₂ = concentration after adding water = 1.75 mol/L and V₂ = volume after adding water = 500 mL = 0.5 L
V₂ = V₁ + V' where V' = volume of water added.
So, From C₁V₁ = C₂V₂
V₁ = C₂V₂/C₁
= 1.75 mol/L × 0.5 L ÷ 8.61 mol/L
= 0.875 mol/8.61 mol/L
= 0.102 L
So, V₂ = V₁ + V'
0.5 L = 0.102 L + V'
V' = 0.5 L - 0.102 L
= 0.398 L
= 398 mL
So, we need to add 398 mL of water to the nitric solution.
Answer:
ΔH = - 272 kJ
Explanation:
We are going to use the fact that Hess law allows us to calculate the enthalpy change of a reaction no matter if the reaction takes place in one step or in several steps. To do this problem we wll add two times the first step to second step as follows:
N2(g) + 3H2(g) → 2NH3(g) ΔH=−92.kJ Multiplying by 2:
2N2(g) + 6H2(g) → 4NH3(g) ΔH=− 184 kK
plus
4NH3(g) + 5O2(g) → 4NO(g) +6H2O(g) ΔH=−905.kJ
__________________________________________________
2N2(g) + 6H2(g) + 5O2(g)→ 4NO(g) + 6H2O(g) ΔH = (-184 +(-905 )) kJ
ΔH = -1089 kJ
Notice how the intermediate NH3 cancels out.
As we can see this equation is for the formation of 4 mol NO, and we are asked to calculate the ΔH for the formation of one mol NO:
-1089 kJ/4 mol NO x 1 mol NO = -272 kJ (rounded to nearest kJ)
NaOH+HCl-> NaCl+H2O
1 mole of NaOH
1 mole of HCl.
To calculate volume of NaOH
CaVa/CbVb= Na/Nb
Where Ca=2M
Cb=1M
Va=200cm³
Vb=xcm³
Substitute into the equation.
2×200/1×Vb=1/1
400/Vb=1/1
Cross multiply
Vb×1=400×1
Vb=400cm³
To calculate the mass of sodium chloride, NaCl from the neutralization rxn.
Mole of NaCl=1
Molar mass of NaCl= 23+35.5=58.5
Mass=xgrammes.
Mass of NaCl=Number of moles × Molar mass.
Substitute
Mass of NaCl= 1×58.5
=58.5g
This is what I could come up with.
Answer:
58.316 is the formula weight of magnesium hydroxide
Answer:
magnesium chloride: MgCl2
zinc sulphide: ZnS
sodium sulphate: Na2SO4
calcium bicarbonate: Ca(HCO3)2
