Moles of Li2CO3 = 1.53/73.891 = 0.0207 mole
Since HCl is in excess, amount of CO2 will depend on the limiting reagent which is Li2CO3.
∴Moles of CO2 = Moles of Li2CO3 = 0.0207.
Answer:
1.0 L
Explanation:
Given that:-
Mass of 
= 
Molar mass of = 64.099 g/mol
The formula for the calculation of moles is shown below:
Thus,
According to the given reaction:-
1 mole of on reaction forms 1 mole of 
0.0396 mole of on reaction forms 0.0396 mole of
Moles of = 0.0396 moles
Considering ideal gas equation as:-
where,
P = pressure of the gas = 742 mmHg
V = Volume of the gas = ?
T = Temperature of the gas = ![26^oC=[26+273]K=299K](https://tex.z-dn.net/?f=26%5EoC%3D%5B26%2B273%5DK%3D299K)
R = Gas constant = 
n = number of moles = 0.0396 moles
Putting values in above equation, we get:
<u>1.0 L of acetylene can be produced from 2.54 g .</u>
Answer:
0.33 mol/kg NH₃
Explanation:
Data:
b(NH₃) = 0.33 mol/kg
b(Na₂SO₄) = 0.10 mol/ kg
Calculations:
The formula for the boiling point elevation ΔTb is
i is the van’t Hoff factor — the number of moles of particles you get from a solute.
(a) For NH₃,
The ammonia is a weak electrolyte, so it exists almost entirely as molecules in solution.
1 mol NH₃ ⟶ 1 mol particles
i ≈ 1, and ib = 1 × 0.33 = 0.33 mol particles per kilogram of water
(b) For Na₂SO₄,
Na₂SO₄(aq) ⟶ 2Na⁺(aq) + 2SO₄²⁻(aq)
1 mol Na₂SO₄ ⟶ 3 mol particles
i = 1 and ib = 3 × 0.10 = 0.30 mol particles per kilogram of water
The NH₃ has more moles of particles, so it has the higher boiling point.
Glucose can be converted into ethanol, carbon dioxide and energy (usually in the form of heat).
Using chemical formulas, this can be illustrated as follows:
Glucose = Ethanol + Carbon Dioxide + Energy
C6H12O6 = 2C2H5OH + 2CO2 + Energy
From the periodic table:
molecular mass of carbon = 12 grams
molecular mass of oxygen = 16 grams
molecular mass of hydrogen = 1 gram
Therefore:
molar mass of glucose = 6(12) + 12(1) + 6(16) = 180 grams
molar mass of ethanol = 2(12) + 5(1) + 16 + 1 = 46 grams
molar mass of carbon dioxide = 1(12) + 2(16) = 44 grams
Based on the conversion equation, each one mole of glucose converts into two moles of ethanol and two moles of carbon dioxide.
Therefore, each 180 grams of glucose converts into 46 x 2 = 92 grams of ethanol and 44 x 2 = 88 grams of carbon dioxide in addition to energy.
To calculate the mass of ethanol and carbon dioxide produced from 200 grams of glucose, we will simply use cross multiplication as follows:
mass of ethanol = (200 x 92) / 180 = 102.2 grams
mass of carbon dioxide = (200 x 88) / 180 = 97.7 grams
Total mass of ethanol and carbon dioxide = 102.2 + 97.7 = 199.9 grams
