Answer: It will occupy 
at the same temperature and 475 mm Hg.
Explanation:
Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.
(At constant temperature and number of moles)
(At constant temperature and number of moles)
where,
= initial pressure of gas = 760 mm Hg
= final pressure of gas = 475 mm Hg
= initial volume of gas = 
= final volume of gas = ?
Putting in the values:
Thus it will occupy at the same temperature and 475 mm Hg
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
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:
100g/mol
Explanation:
Given parameters:
Mass of unknown gas = 2g
Volume of gas in flask = 500mL = 0.5dm³
Unknown:
Molar mass of gas = ?
Solution:
Since we know the gas is at STP;
1 mole of substance occupies 22.4dm³ of space at STP
Therefore,
0.5dm³ will have 0.02mole at STP
Now;
Number of moles = 
Molar mass = 
= 
= 100g/mol
Answer:
The correct answer is is option B
b. 93.3 g
Explanation:
SEE COMPLETE QUESTION BELOW
Hydrogen chloride gas can be prepared by the following reaction: 2NaCl(s) + H2SO4(aq) → 2HCl(g) + Na2SO4(s)
How many grams of HCl can be prepared from 2.00 mol H2SO4 and 2.56 mol NaCl?
a. 7.30 g
b. 93.3 g
c. 146 g
d. 150 g
e. 196 g
CHECK THE ATTACHMENT FOR STEP BY STEP EXPLANATION
