Answer:
H₂ gas
Explanation:
The reaction between nitrogen gas and hydrogen gas forms ammonia (the Haber-Bosch process):
N₂ + 3H₂ ⇒ 2NH₃
The excess reactant can be found by comparing the moles of nitrogen and hydrogen. The molar mass of N₂ is 28.00 g/mol and the molar mass of H₂ is 2.02 g/mol.
(100 kg N₂)(1000g/kg)(mol/28.00g) = 3570 mol
(100 kg H₂)(1000g/kg)(mol/2.02g) = 49500 mol
The molar ratio between the reactant N₂ and H₂ is 1N₂:3H₂. The moles of nitrogen required to react with H₂ is:
(49500 mol H₂)(1N₂ / 3H₂) = 16500 mol
The amount of nitrogen required is more than what is available, so nitrogen is the limiting reagent and hydrogen is the excess reagent.
A solution with a molarity of 7.65 mol/L and a volume of solution of 0.690 L, contains 5.28 moles of solute.
<h3>What is molarity?</h3>
Molarity is the number of moles of solute per liter of solution.
A solution has a molarity (M) of 7.65 mol/L and a volume (V) of 0.690 L. We can calculate the number of moles of solute (n) using the following expression.
M = n / V
n = M × V = 7.65 mol/L × 0.690 L = 5.28 mol
A solution with a molarity of 7.65 mol/L and a volume of solution of 0.690 L, contains 5.28 moles of solute.
Learn more about molarity here: brainly.com/question/26873446
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I think it can be dissolved but if u keep adding more sugar it will stop dissolving
Full electron configuration of barium: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 6s2
Answer:
pH =3.8
Explanation:
Lets call the monoprotic weak acid HA, the dissociation equilibria in water will be:
HA + H₂O ⇄ H₃O⁺ + A⁻ with Ka = [ H₃O⁺] x [A⁻]/ [HA]
The pH is the negative log of the H₃O⁺ concentration, we know the equilibrium constant, Ka and the original acid concentration. So we will need to find the [H₃O⁺] to solve this question.
In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:
HA H₃O⁺ A⁻
Initial, M 0.40 0 0
Change , M -x +x +x
Equilibrium, M 0.40 - x x x
Lets express these concentrations in terms of the equilibrium constant:
Ka = x² / (0.40 - x )
Now the equilibrium constant is so small ( very little dissociation of HA ) that is safe to approximate 0.40 - x to 0.40,
7.3 x 10⁻⁶ = x² / 0.40 ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³
[H₃O⁺] = 1.71 x 10⁻³
Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.
pH = - log ( 1.71 x 10⁻³ ) = 3.8
Note: when the aprroximation is greater than 5 % we will need to solve the resulting quadratic equation.
