Answer:
The mixture contains 8.23 g of Ar
Explanation:
Let's solve this with the Ideal Gases Law
Total pressure of a mixture = (Total moles . R . T) / V
We convert T° from °C to K → 85°C + 273 = 358K
3.43 atm = (Moles . 0.082 L.atm/mol.K . 358K) / 6.47L
(3.43 atm . 6.47L) / (0.082 L.atm/mol.K . 358K) = Moles
0.756= Total moles from the mixture
Moles of Ar + Moles of H₂ = 0.756 moles
Moles of Ar + 1.10 g / 2g/mol = 0.756 moles
Moles of Ar = 0.756 moles - 0.55 moles H₂ → 0.206
We convert the moles to g → 0.206 mol . 39.95 g / 1 mol = 8.23 g
The answer for the following problem is mentioned below.
- <u><em>Therefore number of molecules(N) present in the calcium phosphate sample are 19.3 × 10^23 molecules.</em></u>
Explanation:
Given:
mass of calcium phosphate (
) = 125.3 grams
We know;
molar mass of calcium phosphate (
) = (40×3) + 3 (31 +(4×16))
molar mass of calcium phosphate (
) = 120 + 3(95)
molar mass of calcium phosphate (
) = 120 +285 = 405 grams
<em>We also know;</em>
No of molecules at STP conditions(
) = 6.023 × 10^23 molecules
To solve:
no of molecules present in the sample(N)
We know;
N÷
=
N =(405×6.023 × 10^23) ÷ 125.3
N = 19.3 × 10^23 molecules
<u><em>Therefore number of molecules(N) present in the calcium phosphate sample are 19.3 × 10^23 molecules</em></u>
Answer:
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Explanation:
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