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Flura [38]
3 years ago
14

What volume of gold would be equal in mass to a piece of copper with a volume of 141 ml? the density of gold is 19.3 g/ml; the d

ensity of copper is 8.96 g/ml?
Chemistry
1 answer:
sergejj [24]3 years ago
8 0
We are tasked to solve for the volume of Gold given that the mass of Gold is equivalent to that of a copper.

Details:

Density of Copper= 8.96 g/ml
Volume of Copper=141 ml
mass of Gold = mass of Copper
Density of Gold=19.3 g/ml

In order to solve for the mass of copper, we need to use the density formula
Density= mass/volume

Since we are to solve for the mass of copper
mass of copper= Density of Copper * Volume of Copper
mass of copper= 8.96 g/ml* 141 ml= 1263.36 g
Thus,
mass of gold=mass of copper=1263.36 g
Hence,
Using still the density formula to solve for the volume of gold,
Volume of gold=mass of gold/ Density of gold
Volume of gold=1263.36 g/ 19.3 g/ml = 65.46 mL

Therefore, the volume of the gold must be 65.46 mL in order to have the same mass of a copper
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What are the molality and mole fraction of solute in a 22.3 percent by mass aqueous solution of formic acid (HCOOH)?
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Answer:

Mole fraction for solute = 0.1, or 10%

Molality = 6.24 mol/kg

Explanation:

22.3% by mass → In 100 g of solution, we have  22.3 g of HCOOH

Mass of solution = 100 g

Mass of solute = 22.3 g

Mass of solvent = 100 g - 22.3g = 77.7 g

Let's convert the mass to moles

22.3 g . 1mol/ 46 g = 0.485 moles

77.7 g. 1mol / 18 g = 4.32 moles

Total moles = 4.32 moles + 0.485 moles = 4.805 moles

Xm for solute = 0.485 / 4.805 = 0.100 → 10%

Molality → mol/ kg → we convert the mass of solvent to kg

77.7 g.  1 kg / 1000g = 0.0777 kg

0.485 mol / 0.0777 kg = 6.24 m

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A substance that loses electron and itself is oxidized is called a?​
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a) What substances are present in an aqueous buffer composed of HC2H3O2 and C2H3O2 - ?b) What happens when LiOH is added to a bu
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Answer:

a) HC₂H₃O₂, C₂H₃O₂⁻, H₃O⁺, H₂O, OH⁻

b) HC₂H₃O₂ + LiOH ⇄ H₂O + LiC₂H₃O₂

c) C₂H₃O₂⁻ + HBr ⇄ HC₂H₃O₂ + Br⁻

Explanation:

a) In a HC₂H₃O₂/C₂H₃O₂⁻ buffer system, the following reactions take place:

HC₂H₃O₂ + H₂O ⇄ C₂H₃O₂⁻ + H₃O⁺

C₂H₃O₂⁻ + H₂O ⇄ HC₂H₃O₂ + OH⁻

Thus, the species present are: HC₂H₃O₂, C₂H₃O₂⁻, H₃O⁺, H₂O, OH⁻.

b) When LiOH is added to the buffer system, it is partially neutralized according to the following equation.

HC₂H₃O₂ + LiOH ⇄ H₂O + LiC₂H₃O₂

c) When HBr is added to the buffer system, it is partially neutralized according to the following equation.

C₂H₃O₂⁻ + HBr ⇄ HC₂H₃O₂ + Br⁻

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