Answer:
(a) HCl
(b) HCl
(c) HCl
(d) HCl
Explanation:
<em>Given: </em>0.50 mol of CH₄ and 1.0 mol of HCl
Using stoichiometry we can calculate the answers to parts a, b, c, and d.
<h3>Part (a) </h3>
# of moles × Avogadro's number = # of atoms or molecules
Avogadro's number: 6.02 * 10²³
HCl has more atoms than CH₄.
<h3>Part (b) </h3>
This is calculated the same way as Part (a); HCl has more molecules than CH₄.
<h3>Part (c) </h3>
Molar mass of CH₄ = 16.04 g/mol
Molar mass of HCl = 36.458 g/mol
HCl has a greater mass than CH₄.
<h3>Part (d)</h3>
Assuming STP:
Molar volume of any gas at STP is 22.4 L/mol.
HCl has a greater volume than CH₄.
Answer:
i answered the other one in few
Explanation:
Answer:
Mass released = 8.6 g
Explanation:
Given data:
Initial number of moles nitrogen= 0.950 mol
Initial volume = 25.5 L
Final mass of nitrogen released = ?
Final volume = 17.3 L
Solution:
Formula:
V₁/n₁ = V₂/n₂
25.5 L / 0.950 mol = 17.3 L/n₂
n₂ = 17.3 L× 0.950 mol/25.5 L
n₂ = 16.435 L.mol /25.5 L
n₂ = 0.644 mol
Initial mass of nitrogen:
Mass = number of moles × molar mass
Mass = 0.950 mol × 28 g/mol
Mass = 26.6 g
Final mass of nitrogen:
Mass = number of moles × molar mass
Mass = 0.644 mol × 28 g/mol
Mass = 18.0 g
Mass released = initial mass - final mass
Mass released = 26.6 g - 18.0 g
Mass released = 8.6 g
Answer:
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Explanation:
