Answer:
0.019 moles of M2CO3
Explanation:
M2CO3(aq) + BaCl2 (aq) --> 2MCl (aq) + BaCO3(s)
From the equation above;
1 mol of M2CO3 reacts to produce 1 mol of BaCO3
Mass of BaCO3 formed = 3.7g
Molar mass of BaCO3 = 197.34g/mol
Number of moles = Mass / Molar mass = 3.7 / 197.34 = 0.0187 ≈ 0.019mol
Since 1 mol of M2CO3 reacts with 1 mol of BaCO3,
1 = 1
x = 0.019
x = 0.019 moles of M2CO3
Using ideal gas equation,
P\times V=n\times R\times T
Here,
P denotes pressure
V denotes volume
n denotes number of moles of gas
R denotes gas constant
T denotes temperature
The values at STP will be:
P=100 kPa
T=293 K
R=8.314472 L kPa K⁻¹ mol⁻¹
Number of moles of gas=3.43 mole
Putting all the values in the above equation,
V=83.55 L
So the volume will be 83.55 L.
83.55 L of radon gas would be in 3.43 moles at room temperature and pressure (293 K and 100 kPa).
Answer
D 160g
Explanation:
<u>Write the equation:</u>
Combustion reactions use oxygen and release water and heat, so
CH₃OH(g) + O₂(g) → CO₂(g) + H₂O(g)
Balance that:
2CH₃OH(g) + 3O₂(g) → 2CO₂(g) + 4H₂O(g)
<u>Find moles of carbon dioxide:</u>
We need to know the number of moles of CO₂. This rxn is at STP, so at STP one mole of gas = 22.4 liters.
112 L * 1 mol/22.4 L = <em>5 mol CO₂</em>
<u>Find moles of methanol:</u>
Based on the chemical equation, for every 2 mol methanol, there are 2 mol carbon dioxide. So for every 5 mol carbon dioxide, there are 5 mol methanol!
5 mol CO₂ = 5 mol CH₃OH
Molar mass of methanol: 12.01 + 3*1.008 + 16.00 + 1.008 = <em>32.04 g/mol</em>
Moles of methanol: 5 mol * 32.04 g/mol = 160.2 g methanol
≈ 160 mol methanol
Answer:
The answer to your question is: 1, 2, 1, 2
Explanation:
1 Fe(s) + 2 Na⁺(aq) → 1 Fe²⁺(aq) + 2 Na(s)
Fe⁰ - 2e⁻ ⇒ Fe⁺² Oxidases
Na⁺ + 1 e⁻ ⇒ Na⁰ Reduces
1 x ( 1 Fe⁰ ⇒ 1 Fe⁺²) Interchange number of
2 x ( 2Na⁺ ⇒ 2 Na⁰ ) electrons
