Answer:
Kc = 50.5
Explanation:
We determine the reaction:
H₂ + I₂ ⇄ 2HI
Initially we have 0.001 molesof H₂
and 0.002 moles of I₂
If we have produced 0.00187 moles of HI in the equilibrium we have to know, how many moles of I₂ and H₂, have reacted.
H₂ + I₂ ⇄ 2HI
In: 0.001 0.002 -
R: x x 2x
Eq: 0.001-x 0.002-x 0.00187
x = 0.00187/2 = 9.35×10⁻⁴ moles that have reacted
So in the equilibrium we have:
0.001 - 9.35×10⁻⁴ = 6.5×10⁻⁵ moles of H₂
0.002 - 9.35×10⁻⁴ = 1.065×10⁻³ moles of I₂
Expression for Kc is = (HI)² / (H₂) . (I₂)
0.00187 ² / 6.5×10⁻⁵ . 1.065×10⁻³ = 50.5
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Answer:
Explanation:
Hello,
In this case, since nitric acid is HNO₃ and strontium hydroxide is Sr(OH)₂ we can represent the balanced chemical reaction by equaling the atoms of strontium, nitrogen, oxygen and hydrogen at both reactants and products as shown below:
Best regards.
Answer:
The molar mass of NaOH is 40.00 g/mol.
The number of moles of NaOH is equal to the ratio of its mass to molar mass.
The number of moles of NaOH =
40 g/mol
40 g
=1,000 mol
The molarity of NaOH solution is the ratio of number of moles of NaOH to total volume of solution in L.
M=
1 L
1 mol
=1 M
Explanation ;)
<u>Given:</u>
Mass of Na = 115 g
Excess Cl2
<u>To determine:</u>
Mass of NaCl produced
<u>Explanation:</u>
Given reaction is-
2Na(s) + Cl2(g) → 2NaCl(s)
Since Cl2 is in excess, Na will be the limiting reagent
As per the reaction stoichiometry Na:NaCl = 1:1
i.e. moles of Na reacted = moles of NaCl formed
Now, # moles of Na = mass of Na/atomic mass
= 115 g/23 g.mol-1 = 5 moles
Therefore, moles of NaCl = 5
Molar mass of NaCl = 58 g/mol
Mass of NaCl = 5 moles * 58 g.mol-1 = 290 g
Ans: Amount of Nacl produced = 290 g
