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
Answer:
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Answer:
Mass in kg = 4.7*10^19 kg
Mass in tons = 5.2*10^16 tons
Explanation:
<u>Given:</u>
Total volume of sea water = 1.5*10^21 L
Mass % NaCl in seawater = 3.1%
Density of seawater = 1.03 g/ml
<u>To determine:</u>
Total mass of NaCl in kg and in tons
<u>Calculation:</u>
Unit conversion:
1 L = 1000 ml
The volume of seawater in ml is:



To convert mass from g to Kg:
1000 g = 1 kg

To convert mass from g to tons:
1 ton = 9.072*10^6 g
