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Aleksandr-060686 [28]
3 years ago
9

A sample of aluminum is placed in a 25 ml graduated cylinder containing 10 mL of water. The level of water rises to 18 mL. Alumi

num has a density of 2.7 g/mL. Calculate the mass of the sample
Please include explanation if possible
Chemistry
2 answers:
Readme [11.4K]3 years ago
8 0

Answer: The mass of sample is 21.6 grams.

Explanation:

We are given:

Volume of cylinder without object, V_1 = 10 mL

Volume of cylinder with object, V_2 = 18 mL

Volume of object = V_2-V_1=18-10=8mL

To calculate volume of a substance, we use the equation:

\text{Density of a substance}=\frac{\text{Mass of a substance}}{\text{Volume of a substance}}

We are given:

Density of sample (aluminum) = 2.7 g/mL

Volume of sample (aluminum) = 8 mL

Putting values in above equation, we get:

2.7g/mL=\frac{\text{Mass of a sample}}{8mL}\\\\\text{Mass of object}=21.6g

Hence, the mass of sample is 21.6 grams.

anastassius [24]3 years ago
4 0

To determine the mass of the sample, first find the volume difference after and before the aluminum was placed, the volume change is equal to the volume of the submerged object, in this case aluminum.

Then knowing volume of aluminum and the density of it, we can solve for the mass.

D = m/v

Dv = m

2.7 g/ml • 8 ml = 21.6 grams.

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Predict which of the following pairs of solutions, when mixed together, will cause a precipitate to form. (Select all that apply
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Answer:

25 mL of 1 × 10–5 M Co(NO₃)₂ and 75 mL of 5 × 10–4 M Na₂S

500 mL of 7.5 × 10–4 M AlCl₃ and 100 mL of 1.7 × 10–5 M Hg₂(NO₃)₂.

650 mL of 0.0080 M K₂SO₄ and 175 mL of 0.15 M AgNO₃

Explanation:

When 2 compounds that produce an insoluble substance are mixed together, <em>A precipitate will be formed if Q of reaction > Ksp</em>

For the solutions:

1.5 L of 0.025 M BaCl₂ and 1.25L of 0.014 M Pb(NO₃)₂.

Ksp is:

PbCl₂(s) ⇄ Pb²⁺(aq) + 2Cl⁻(aq)

Ksp = 2.4x10⁻⁴ = [Pb²⁺][Cl⁻]²

Molar concentration of each ion is:

[Pb²⁺] =  1.25L ₓ (0.014mol / L) = 0.0175mol / 2.75L = 6.36x10⁻³M

[Cl⁻] = 2 ₓ 1.5L ₓ (0.025mol / L) = 0.075mol / 2.75L = 0.0273M

Replacing in Ksp expression to find Q:

Q = [6.36x10⁻³M][0.0273M]² = 4.73x10⁻⁶

As Q < Ksp, the mixture will not produce a precipitate.

25 mL of 1 × 10–5 M Co(NO₃)₂ and 75 mL of 5 × 10–4 M Na₂S

Ksp is:

CoS(s) ⇄ Co²⁺(aq) + S²⁻(aq)

Ksp = 4.0x10⁻²¹ = [Co²⁺][S²⁻]

Molar concentration of each ion is:

[Co²⁺] =  0.025L ₓ (1x10⁻⁵mol / L) = 2.5x10⁻⁷mol / 0.1L = 2.5x10⁻⁶M

[S²⁻] = 0.075L ₓ (5x10⁻⁴mol / L) = 3.75x10⁻⁵mol / 0.1L = 3.75x10⁻⁴M

Replacing in Ksp expression to find Q:

Q = [2.5x10⁻⁶M][3.75x10⁻⁴M] = 9.38x10⁻⁶

As Q > Ksp, the mixture will produce a precipitate.

500 mL of 7.5 × 10–4 M AlCl₃ and 100 mL of 1.7 × 10–5 M Hg₂(NO₃)₂.

Ksp is:

Hg₂Cl₂(s) ⇄ 2Hg⁺(aq) + 2Cl⁻(aq)

Ksp = 3.5x10⁻¹⁸ = [Hg⁺]²[Cl⁻]²

Molar concentration of each ion is:

[Hg⁺] =  2ₓ0.100L ₓ (1.7x10⁻⁵mol / L) = 3.4x10⁻⁶mol / 0.6L = 5.67x10⁻⁶M

[Cl⁻] = 3 ₓ 0.500L ₓ (7.5x10⁻⁴mol / L) = 1.125x10⁻³mol / 0.6L = 1.88x10⁻³M

Replacing in Ksp expression to find Q:

Q = [5.67x10⁻⁶M]²[1.88x10⁻³M]² = 1.14x10⁻⁶

As Q > Ksp, the reaction will produce a precipitate.

650 mL of 0.0080 M K₂SO₄ and 175 mL of 0.15 M AgNO₃

Ksp is:

Ag₂SO₄(s) ⇄ 2Ag⁺(aq) + SO₄²⁻(aq)

Ksp = 1.5x10⁻⁵ = [Ag⁺]²[SO₄²⁻]

Molar concentration of each ion is:

[Ag⁺] =  0.175L ₓ (0.15mol / L) = 0.02625mol / 0.825L = 0.0318M

[SO₄²⁻] = 0.650L ₓ (0.080mol / L) = 0.052mol / 0.825L = 0.0630M

Replacing in Ksp expression to find Q:

Q = [0.0318M]²[0.0630M] = 6.37x10⁻⁵

As Q > Ksp, the reaction will produce a precipitate.

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In the first 22.0 s of this reaction, the concentration of HBr dropped from 0.590 M to 0.465 M . Calculate the average rate of t
jolli1 [7]

Answer:

The average rate is 2.84 X 10⁻³ Ms⁻¹

Explanation:

Average rate = -0.5*Δ[HBr]/Δt

given;

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[HBr]₂ = 0.465 M

Δ[HBr] = [HBr]₂  - [HBr]₁ = 0.465 M - 0.590 M = -0.125 M

Δt Change in time = 22.0 s

Average rate = -0.5*Δ[HBr]/Δt

Average rate = - 0.5(-0.125)/22

Average rate = 0.00284 Ms⁻¹ = 2.84 X 10⁻³ Ms⁻¹

Therefore, the average rate is 2.84 X 10⁻³ Ms⁻¹

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