Answer:
62.98 % of the sample of hydrate is water
Explanation:
Step 1: Data given
Mass of the sample of a hydrate of sodium carbonate (Na2CO3) = 2.026 grams
After heating, the mass of the sample is 0.750 g
Molar mass H2O = 18.02 g/mol
Step 2: Calculate mass of water
Mass water = mass of hydrate - mass of sample after heating
Mass water = 2.026 grams - 0.750 grams
Mass water = 1.276 grams
Step 3: Calculate mass % percent of water
Mass % of water = (mass of water / total mass hydrate) * 100 %
Mass % of water = (1.276 grams / 2.026 grams) *100 %
Mass % of water = 62.98 %
62.98 % of the sample of hydrate is water
Answer: 600 kJ
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Explanation:
C₃H₈ (g) + 5 O₂ (g) =============== 3 CO₂ (g) + 4 H₂O (l)
Δ⁰Hf kJ/mol -104 0 -393.5 -285.8
Δ⁰Hcomb C₃H₈ = 3(-393.5) + 4 (-285.80) - (-104) kJ/mol
Δ⁰Hcomb = 2219.70 kJ/mol
n= m /MW MW c₃H₈ = 44.1 g/mol
n= 12 g/44.1 g/mol = 0.27 mol
then for 12 g the heat released will be
0.27 mol x 2219.70 kJ/mol = 600 KJ
Answer:
....,................................
Explanation:
1= A
2=D
3=C
4=C
Answer:
Lithium Fluoride
Explanation:
It should be lithium fluoride for sure :)
