Ammonium hydrogen sulfide NH4HS(s) decomposes on heating according to the reaction NH4HS(s) ↔ NH3(g) + H2S(g) At 25 °C the equil
2 answers:
Answer:
0.66atm
Explanation:
Based on the reaction:
NH₄HS(s) ↔ NH₃(g) + H₂S(g)
The equilibrium constant, Kp, is defined as:
Kp = 0.11 =
As moles of gas produced for NH₃(g) and H₂S(g) are the same, it is possible to write:
That means pressure of NH₃(g) is 0.33atm and H₂S(g) is, also, 0.33atm. Thus, total pressure is:
0.33atm×2 = <em>0.66atm</em>
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Answer:
28 g CO
Explanation:
First convert grams to moles.
1 mole C = 12.011 g (I'm just going to round to 12 for the sake of this problem)
12 g C • = 1 mol C
1 mol O = 15.996 g (I'm just going to round to 16)
16 g O • = 1 mol O
So the unbalanced equation is:
->
(the oxygen has a 2 subscript because it is part of HONClBrIF meaning when not in a compound these elements appear in pairs - called diatomic elements)
The balanced equation is:
->
However, carbon is the limiting reactant in this equation and two moles cannot react because only 12 g (1 mole) are present. Therefore, use the equation
->
.
1 mole of CO is formed, therefore 12 g + 16 g = 28 g CO.
Answer:
In a titration of 35.00 mL of 0.737 M H₂SO₄, 62.4 mL of a 0.827 M KOH solution is required for neutralization.
Explanation:
The balanced reaction is
H₂SO₄ + 2 KOH ⇒ 2 H₂O + K₂SO₄
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) 1 mole of H₂SO₄ is neutralized with 2 moles of KOH.
The molarity M being the number of moles of solute that are dissolved in a given volume, expressed as:
in units of
then the number of moles can be calculated as:
number of moles= molarity* volume
You have acid H₂SO₄
- 35.00 mL= 0.035 L (being 1,000 mL= 1 L)
- Molarity= 0.737 M
Then:
number of moles= 0.737 M* 0.035 L
number of moles= 0.0258
So you must neutralize 0.0258 moles of H₂SO₄. Now you can apply the following rule of three: if by stoichiometry 1 mole of H₂SO₄ are neutralized with 2 moles of KOH, 0.0258 moles of H₂SO₄ are neutralized with how many moles of KOH?
moles of KOH= 0.0516
Then 0.0516 moles of KOH are needed. So you know:
- Molarity= 0.827 M
- number of moles= 0.0516
- volume=?
Replacing in the definition of molarity:
Solving:
volume=0.0624 L= 62.4 mL
<u><em>In a titration of 35.00 mL of 0.737 M H₂SO₄, 62.4 mL of a 0.827 M KOH solution is required for neutralization.</em></u>