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marishachu [46]
3 years ago
8

The density of gold is 19.3g/mL. If you have a 9.00 g piece of gold, what volume will it occupy (in milliliters)

Chemistry
1 answer:
kiruha [24]3 years ago
5 0

Answer:

The answer to your question is volume = 0.47 ml

Explanation:

To solve this problem use the formula of density, that relates the mass and volume.

Data

density = 19.3 g/ml

mass = 9.0 g

volume = ?

Formula

density = \frac{mass}{volume}

Solve for volume

volume = \frac{mass}{density}

Substitution

Volume = 9 / 19,3

Simplification and result

Volume = 0.47 ml

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When blood is donated, sodium oxalate solution is used to precipitate Ca2+, which triggers clotting. A 122.0−mL sample of blood
icang [17]

Answer:

[Ca2+] = 3.36 * 10^-8 M

Explanation:

Step 1: Data given

A 122.0−mL sample of blood contains 9.70 * 10^−5 g Ca2+/mL.

A technologist treats the sample with 100.0 mL of 0.1550 M Na2C2O4.

Ksp = 2.30 * 10^−9

Step 2: Calculate mass of Ca2+

Mass Ca2+ = 9.70 * 10^−5 g Ca2+/mL * 122 mL

Mass Ca2+ = 0.011834 grams

Step 3: Calculate moles of Ca2+

Moles Ca2+ = mass Ca2+ / molar mass Ca2+

Moles Ca2+ = 0.011834 grams / 40.078 g/mol

Moles Ca2+ = 2.95 *10^-4 moles = 0.000295

Step 4: Calculate moles of C2O4^2-

Moles C2O4^2- = Molarity * volume

Moles C2O4^2- = 0.1550 M * 0.100 L

Moles C2O4^2- = 0.0155 moles

Step 5: Calculate limiting reactant

Ca2+ is the limiting reactant. It will completely be consumed ( 0.000295 moles).

C2O4^2- is in excess. There will 0.000295 moles be consumed. There will remain 0.0155 - 0.000295 = 0.015205 moles of C2O4^2-

Step 6: Calculate total volume

Total volume = 122.0 mL + 100.0 mL = 222.0 mL = 0.222 L

Step 7: Calculate concentration CaC2O4

[CaC2O4] = 0.000295 mol / 0.222 L

[CaC2O4] = 0.00133 M

Step 8: Calculate concentration of C2O4^2-

[C2O4^2-] = 0.015205 mol / 0.222L

[C2O4^2-] = 0.0685 M

Step 9: Calculate [Ca2+]

Ksp = 2.3 * 10^-9 = [Ca2+] [C2O42-]

2.3 * 10^-9 = X * (X+0.0685)

X = [Ca2+] = 3.36 * 10^-8 M

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