Answer:
5×10⁵ L of ammonia (NH3)
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
N2 + 3H2 —> 2NH3
From the balanced equation above, we can say that:
3 L of H2 reacted to produce 2 L of NH3.
Finally, we shall determine the volume of ammonia (NH3) produced by the reaction of 7.5×10⁵ L of H2. This can be obtained as illustrated below:
From the balanced equation above,
3 L of H2 reacted to produce 2 L of NH3.
Therefore, 7.5×10⁵ L of H2 will react to produce = (7.5×10⁵ × 2)/3 = 5×10⁵ L of NH3.
Thus, 5×10⁵ L of ammonia (NH3) is produced from the reaction.
Answer:
92.49 %
Explanation:
We first calculate the number of moles n of AgBr in 0.7127 g
n = m/M where M = molar mass of AgBr = 187.77 g/mol and m = mass of AgBr formed = 0.7127 g
n = m/M = 0.7127g/187.77 g/mol = 0.0038 mol
Since 1 mol of Bromide ion Br⁻ forms 1 mol AgBr, number of moles of Br⁻ formed = 0.0038 mol and
From n = m/M
m = nM . Where m = mass of Bromide ion precipitate and M = Molar mass of Bromine = 79.904 g/mol
m = 0.0038 mol × 79.904 g/mol = 0.3036 g
% Br in compound = m₁/m₂ × 100%
m₁ = mass of Br in compound = m = 0.3036 g (Since the same amount of Br in the compound is the same amount in the precipitate.)
m₂ = mass of compound = 0.3283 g
% Br in compound = m₁/m₂ × 100% = 0.3036/0.3283 × 100% = 0.9249 × 100% = 92.49 %
Answer:
Mass = 64 g
Explanation:
Given data:
Mass of water produced = 36 g
Mass of oxygen needed = ?
Solution:
Chemical equation:
CH₄ + 2O₂ CO₂ + 2H₂O
Number of moles of water produced:
Number of moles = mass/molar mass
Number of moles = 36 g/ 18 g/mol
Number of moles = 2 mol
Now we will compare the moles of water and oxygen.
H₂O : O₂
2 : 2
Mass of oxygen:
Mass = number of moles × molar mass
Mass = 2 mol × 32 g/mol
Mass = 64 g
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1. would be deposition
2.sublimation
