Answer:
Volume of ammonia produced = 398.7 dm³
Explanation:
Given data:
Volume of N₂ = 200 dm³
Pressure and temperature = standard
Volume of ammonia produced = ?
Solution:
Chemical equation:
N₂ + 3H₂ → 2NH₃
Number of moles of N₂:
PV = nRT
1 atm× 200 L = n× 0.0821 atm.L/mol.K × 273 K
n = 200 atm.L /22.41 atm.L/mol
n = 8.9 mol
Now we will compare the moles of ammonia and nitrogen.
N₂ : NH₃
1 : 2
8.9 : 2/1×8.9 = 17.8 mol
Volume of ammonia:
1 mole of any gas occupy 22.4 dm³ volume
17.8 mol ×22.4 dm³/1 mol = 398.7 dm³
Answer:
'See Explanation
Explanation:
Determine the [OH−] , pH, and pOH of a solution with a [H+] of 9.5×10−13 M at 25 °C.
Given [H⁺] = 9.5 x 10⁻¹³M => [H⁺][OH⁻] = 1.0 x 10⁻¹⁴ => [OH⁻] = 1.0 x 10⁻¹⁴/9.5 x 10⁻¹³ = 0.0105M
pH = -log[H⁺] = -log(9.5 x 10⁻¹³) = - (-1202) = 12.02.
pOH = -log[OH⁻] = -log(0.0105) = -(-1.98) = 1.98
Now you use the same sequence in the remaining problems.
Answer:
B
Explanation:
Molarity = 0.010M
Volume = 2.5L
Applying mole-concept,
0.010mole = 1L
X mole = 2.5L
X = (0.010 × 2.5) / 1
X = 0.025moles
0.025moles is present in 2.5L of NaOH solution.
Molar mass of NaOH = (23 + 16 + 1) = 40g/mol
Number of moles = mass / molar mass
Mass = number of moles × molar mass
Mass = 0.025 × 40
Mass = 1g
1g is present in 2.5L of NaOH solution
Answer:
Mass = 88.12 g
Explanation:
Given data:
Mass of iron oxide = 126 g
Mass of iron formed = ?
Solution:
Chemical equation:
Fe₂O₃ + 3CO → 2Fe + 3CO₂
Number of moles of iron oxide:
Number of moles = mass/molar mass
Number of moles = 126 g/ 159.69 g/mol
Number of moles = 0.789 mol
Now we will compare the moles of iron with iron oxide.
Fe₂O₃ : Fe
1 : 2
0.789 : 2/1×0.789 = 1.578 mol
Mass of iron:
Mass = number of moles ×molar mass
Mass = 1.578 mol × 55.84 g/mol
Mass = 88.12 g
