How many litres of 0.435 M KBr solution are needed to provide 0.717 mol of KBr?
How much water (mL) must be added to a 0.723 M solution of lithium carbonate with a volume of 17.7 mL so that the final concentration is 0.138 M?
How much energy is required to heat 44.72 g of H2O from 10.00 oC to 37.50 oC? (Heat capacity of liquid H2O = 75.37 J mol-1 K-1) (use molar mass H2O = 18.015 g/mol
quick chem Q's - easy points !?
2020-10-05 4:40 pm
回答 (1)
2020-10-05 7:11 pm
1.
Volume of KBr = (0.717 mol) × (1 L / 0.435 mol) = 1.65 L
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2.
For dilution: M₁V₁ = M₂V₂
Final volume, V₂ = V₁ × (M₁/M₂) = (17.7 mL) × (0.723/0.138) = 92.7 mL
Volume of water added = (92.7 - 17.7) mL = 75.0 mL
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3.
Energy required
= (75.37 J mol⁻¹ K⁻¹) × (44.72/18.015 mol) × [(37.50 - 10.00)°C]
= 5145 J
= 5.145 kJ
Volume of KBr = (0.717 mol) × (1 L / 0.435 mol) = 1.65 L
====
2.
For dilution: M₁V₁ = M₂V₂
Final volume, V₂ = V₁ × (M₁/M₂) = (17.7 mL) × (0.723/0.138) = 92.7 mL
Volume of water added = (92.7 - 17.7) mL = 75.0 mL
====
3.
Energy required
= (75.37 J mol⁻¹ K⁻¹) × (44.72/18.015 mol) × [(37.50 - 10.00)°C]
= 5145 J
= 5.145 kJ
收錄日期: 2021-05-01 22:40:04
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20201005084046AANszOA