Answer:
CH2.
Explanation:
The mass of C in the CO2 = 2.49 * 12/(12+32) = 0.6791 g
The mass of Hydrogen in the water = 2 * 1.008 / (2.016 + 16) = 0.1119 g.
Ratio of H to O is 0.1119/ 1.008 : 0.6791/12
= 0.111 : 0.0566
= 2 : 1
The empirical formula is CH2
The percent yield of the reaction between ammonia gas with oxygen gas is 90.52%.
A chemical reaction between ammonia gas (NH3) with oxygen gas (O2)
NH₃ + O₂ → NO₂ + H₂O
The balanced reaction 4NH₃ + 7O₂ → 4NO₂ + 6H₂O
Calculate the number of moles from the reactant
- Ammonia gas
Molar mass N = 14 gr/mol
Molar mass H = 1 gr/mol
Molar mass NH₃ = 14 + (3 × 1) = 14 + 3 = 17 gr/mol
mass = 28.5 grams
n = m ÷ molar mass = 28.5 ÷ 17 = 1.68 mol - Oxygen gas
Molar mass O = 16 gr/mol
Molar mass O₂ = 16 × 2 = 32 gr/mol
mass = 83.4 grams
n = m ÷ molar mass = 83.4 ÷ 32 = 2.61 mol - n O₂ ÷ coefficient O₂ = 2.61 ÷ 7 = 0.37
n NH₃ ÷ coefficient NH₃ = 1.68 ÷ 4 = 0.42
0.42 > 0.37 it means that the ammonia gas is in excess and the O₂ is limiting.
According to stoichiometry, the number of moles NO₂ with the number of moles O₂ has the ratio with the coefficient in reaction.
- Theoretically the number moles of NO₂
n O₂ : n NO₂ = 7 : 4
2.61 : n NO₂ = 7 : 4
n NO₂ = 4 x 2.61 : 7 = 1.49 mol - The actual number of moles NO₂
Molar mas NO₂ = 14 + (16 × 2) = 14 + 32 = 46 gr/mol
n NO₂ = m ÷ molar mass = 61.9 ÷ 46 = 1.35 mol
The percent yield NO₂ is the ratio of the actual number of moles NO₂ with the theoretical number of moles NO₂ times 100%.
P = (1.35 ÷ 1.49) × 100%
P = 0.9052 × 100%
P = 90.52%
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Answer:
17202.6 years
Explanation:
Activity of the living sample (Ao) = 160 counts per minute
Activity of the wood sample (A) = 20 counts per minute
Half life of carbon-14 = 5730 years
t= age of the artifact
From;
0.693/t1/2= 2.303/t log Ao/A
Then;
0.693/ 5730= 2.303/t log Ao/A
Substituting values;
0.693/5730= 2.303/t log (160/20)
Then we obtain;
1.209×10^-4 = 2.0798/t
t= 2.0798/1.209×10^-4
Thus;
t= 17202.6 years
Therefore the artifact is 17202.6 years old.
Answer: should be 4, 5, 4, 6
Explanation:
CaO + H2O → Ca(OH)2
the number of O atom on the left side is 1 and on the right side 2 so,
we put 2H2O instead of H2O
CaO + 2H2O → Ca(OH)2
now we have to put 2Ca(OH)2 instead of Ca(OH)2 to make H atoms equal on both sides 4
CaO + 2H2O → 2Ca(OH)2
now we have to put 2CaO instead of CaO to make Ca atoms equal on both sides 2 atoms.
∴ the final balanced equation is:
2CaO + 2H2O → 2 Ca(OH)2
∴ the coefficients are 2 2 2