This reaction would produce salt and water- Sodium Sulphate and Water.
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
Answer:
Molar concentration of CO₂ in equilibrium is 0.17996M
Explanation:
Based on the reaction:
NiO(s) + CO(g) ⇆ Ni(s) + CO₂(g)
kc is defined as:
kc = [CO₂] / [CO] = 4.0x10³ <em>(1)</em>
As initial concentration of CO is 0.18M, the concentrations in equilibrium are:
[CO] = 0.18000M - x
[CO₂] = x
Replacing in (1):
4.0x10³ = x / (0.18000-x)
720 - 4000x = x
720 = 4001x
x = 0.17996
Thus, concentrations in equilibrium are:
[CO] = 0.18000M - 0.17996 = 4.0x10⁻⁵
[CO₂] = x = <em>0.17996M</em>
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Thus, <em>molar concentration of CO₂ in equilibrium is 0.17996M</em>
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I hope it helps!
Answer:
259.497 mg, 58.84%
Explanation:
BaSO₄ → Ba²⁺ + SO₄²⁻
to calculate the mole of BaSO₄
mole BaSO₄ = mass given / molar mass = 403 mg / 233.38 g/mol = 1.7268 mol
comparing the mole ratio
1.7268 mol of BaSO₄ yields 1.7268 mol of Ba²⁺
403 mg BaSO₄ yields ( 1.7268 × 137.327 ) where 137.327 is the molar mass of Barium mol of Ba²⁺
441 mg BaSO₄ will yield ( 1.7268 × 137.327 × 441 mg ) / 403 mg = 259 .497 mg
mas percentage of the Barium compound = 259 .497 mg / 441 mg × 100 = 58.84%
Answer:
5 mL
Explanation:
As this is a problem regarding <em>dilutions</em>, we can solve it using the following formula:
Where subscript 1 refers to the initial concentration and volume, while 2 refers to the final C and V. Meaning that in this case:
We <u>input the data</u>:
- 5.0 M * V₁ = 0.25 M * 100 mL
And <u>solve for V₁</u>:
It’s blurry sorry. I’m not sure
