Answer: 0.125moles
Explanation:
From the question above, we are only given the values of the volume and molar concentration of CuSO4 solution.
Volume of CuSO4 solution = 250mL= 250/1000L = 0.25L
Molar concentration of CuSO4 solution= 0.50M
Number of moles of CuSO4= Molar concentration of CuSO4 × volume of CuSO4
Number of moles of CuSO4= 0.50mol/dm^3 × 0.25L
Number of moles of CuSO4= 0.125moles
Therefore 0.125moles of CuSO4 are used to make 250 mL of a 0.50M CuSO4 solution.
Answer:
(a) 1.95 × 10⁴ g
(b) 1.95 × 10⁷ mg
(c) 1.95 × 10¹⁰ μg
Explanation:
A dog has a mass of 19.5 kg.
<em>(a) What is the dog's mass in grams?</em>
1 kilogram is equal to 10³ grams. The mass of the dog in grams is:
19.5 kg × (10³ g/1 kg) = 1.95 × 10⁴ g
<em>(b) What is the dog's mass in milligrams?</em>
1 gram is equal to 10³ milligrams. The mass of the dog in milligrams is:
1.95 × 10⁴ g × (10³ mg/1 g) = 1.95 × 10⁷ mg
<em>(c) What is the dog's mass in micrograms?</em>
1 gram is equal to 10⁶ micrograms. The mass of the dog in micrograms is:
1.95 × 10⁴ g × (10⁶ μg/1 g) = 1.95 × 10¹⁰ μg
the equilibrium concentration of H₂(g) at 700°C = 0.00193 mol/L
0.00193 mol/L
Given that:
numbers of moles of H₂S = 0.59 moles
Volume = 3.0-L
Equilibrium constant = 9.30 × 10⁻⁸
The equation for the reaction is given as :
2H₂S ⇄ 2H₂(g) + S₂(g)
The initial concentration of H₂S =
The initial concentration of H₂S =
= 0.1966 mol/L
The ICE table is shown be as :
2H₂S ⇄ 2H₂(g) + S₂(g)
Initial 0.9166 0 0
Change -2 x +2 x + x
Equilibrium (0.9166 - 2x) 2x x
(since 2x < 0.1966 if solved through quadratic equation)
The equilibrium concentration for H₂(g) = 2x
∴
= 0.00193 mol/L
Thus, the equilibrium concentration of H₂(g) at 700°C = 0.00193 mol/L
To know more about equilibrium concentration
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Answer:
n₂ = 31 mol
Explanation:
Given data:
Initial volume of gas = 6.2 L
Initial number of moles of gas = 19.2 mol
Final number of moles = ?
Final volume of gas = 10.0 L
Solution:
Formula:
V₁/n₁ = V₂/n₂
V₁ = Initial volume
n₁ = Initial number of moles
V₂ = Final volume
n₂ = Final number of mole
Now we will put the values.
6.2 L /19.2 mol = 10.0 L / n₂
n₂ = 10.0 L× 19.2 mol /6.2 L
n₂ = 192 L.mol /6.2 L
n₂ = 31 mol