The problem you have written you almost have it solved. Take the moles that you have calculated and multiply that by the molecular weight to get the grams.
The STP problem:
use the moles you calculated along with 1 atm for Pressure, and 273 for the temperature and plug into the PV = nRT equation. (also use 0.0821 for R)
From there you can solve for the volume
Hope this helps!
Answer:
The molality of isoborneol in camphor is 0.53 mol/kg.
Explanation:
Melting point of pure camphor= T =179°C
Melting point of sample = 
= 165°C
Depression in freezing point = 
Depression in freezing point is also given by formula:
= The freezing point depression constant
m = molality of the sample
i = van't Hoff factor
We have: = 40°C kg/mol
i = 1 ( organic compounds)
The molality of isoborneol in camphor is 0.53 mol/kg.
The volume (in mL) of calcium hydroxide, Ca(OH)₂ needed for the reaction is 19.8 mL
<h3>Balanced equation </h3>
2HCl + Ca(OH)₂ —> CaCl₂ + 2H₂O
From the balanced equation above,
- The mole ratio of the acid, HCl (nA) = 2
- The mole ratio of the base, Ca(OH)₂ (nB) = 1
<h3>How to determine the volume of Ca(OH)₂ </h3>
- Molarity of base, Ca(OH)₂ (Mb) = 1.48 M
- Volume of acid, HCl (Va) = 36 mL
- Molarity of acid, HCl (Ma) = 1.63 M
- Volume of base, Ca(OH)₂ (Vb) =?
MaVa / MbVb = nA / nB
(1.63 × 36) / (1.48 × Vb) = 2
58.68 / (1.48 × Vb) = 2
Cross multiply
2 × 1.48 × Vb = 58.68
2.96 × Vb = 58.68
Divide both side by 2.96
Vb = 58.68 / 2.96
Vb = 19.8 mL
Learn more about titration:
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It’s one... electron affinity
Answer:
According to conservation of matter, there should be equal amounts of all elements on both the reactant and product side.
Reactant:
1 Ca
1 C
1 O
Product:
1 Ca
1 C
3 O
Therefore, your friend is right because the law of conservation of matter is not followed in this chemical equation.
Explanation:
