Probably the answer will be 28.60 because it's a lower number
Answer:
[CH₃COOH] = 17.4 M
Explanation:
Acetic acid → CH₃COOH
Molar mass → 60 g/mol
99.5% is percent by mass concentration. It means that 99.5 grams of solute are contained in 100g of solution.
Density → 1.05 g/mL. This data is always referred to solution, not solute!.
We determine solution's volume:
1.05 g/mL = 100 g / V → V = 100 g /1.05 g/mL → 95.2 mL
Now we know, that 99.5 g of acetic acid are contained in 95.2 mL
Let's convert to mmoles → 99.5 g / 60 g/mol = 1.66 moles
We convert solution's volume to L → 95.2 mL . 1L / 1000 mL = 0.0952 L
M (mol/L) = 1.66 mol / 0.0952 L = 17.4 M
Answer:
At ambient pressure and temperature the isobaric specific heat, CP, of gaseous propane is 1.68 kJ/kg K or 0.402 Btu/lb °F = cal/g K, while the isochoric specific heat, CV, is 1.48 kJ/kg K or 0.353 Btu/lb °F = cal/g K.
Explanation:
Hope this helps
<em>Answer:</em>
- The molarity of ammonia will be 2.88 M.
<em>Chemical equation</em>
HCl + NH3 ------> NH4Cl
First of calculate the moles of HCl
mole of HCl = Molarity × Vol (L)
mole of HCl = 0.800× 0.018 = 0.014 mole
As the in balance chemical, moles of HCl and NH3 areequal
so
moles of NH3= 0.014
Molarity of NH3 = moles ÷ V(L) = 0.014/0.005 = 2.88 M
<em>Result</em>:
- The molarity of ammonia will be 2.88 M.
