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inna [77]
3 years ago
15

This is a steel penny from 1943. A steel penny will weigh 2.72 g and is

Chemistry
1 answer:
ehidna [41]3 years ago
5 0

Answer:

The easiest way to determine if a 1943 cent is made of steel, and not copper, is to use a magnet. If it sticks to the magnet, it is not copper. If it does not stick, the coin might be of copper and should be authenticated by an expert

Explanation:

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Which property is shared between liquids that are used in thermometers.
miv72 [106K]

Answer:

the property that is shared between liquids us called thermometric property your welcome

8 0
2 years ago
Elements symbol
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8 0
3 years ago
Who wants a metal?
tiny-mole [99]

Which best compares 1 mol of sodium chloride to 1 mol of aluminum chloride?

A. Both have the same molar mass.

B. Both have the same number of ions.

C. Both are made up of 6.02 mc014-1 1023 molecules.

D. Both are made up of 6.02 mc014-2 1023 formula units.

The correct answer on E.D.G is ---Both are made up of 6.02 mc014-2 1023 formula units. D

8 0
4 years ago
Read 2 more answers
Predict which of the following pairs of solutions, when mixed together, will cause a precipitate to form. (Select all that apply
Kay [80]

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.

6 0
3 years ago
How many kilograms of oxygen did Samantha need to utilize to react with 36.29 kg of triglycerides (C55H104O6)?
Paraphin [41]

Answer:

105.33Kg of O2.

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

C55H104O6 + 78O2 —> 55CO2 + 52H2O

Step 2:

Determination of the masses of C55H104O6 and O2 that reacted from the balanced equation.

This is illustrated below:

Molar mass of C55H104O6 = (12x55) + (104x1) + (16x6) = 860g/mol

Mass of C55H104O6 from the balanced equation = 1 x 860 = 860g.

Divide the mass of C55H104O6 by 1000 to express in kg i.e

Mass of C55H104O6 = 860/1000= 0.86Kg

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 78 x 32 = 2496g.

Divide the mass of O2 by 1000 to express in kg i.e

Mass of O2 = 2496/1000 = 2.496kg.

From the balanced equation above,

0.86kg of C55H104O6 reacted with 2.496Kg of O2.

Step 3:

Determination of the mass of O2 in kg needed to react with 36.29 kg of triglycerides (C55H104O6).

This can be obtained as follow:

From the balanced equation above,

0.86kg of C55H104O6 reacted with 2.496Kg of O2.

Therefore, 36.29 kg of C55H104O6 will react with = (36.29 x 2.496)/0.86 = 105.33Kg of O2.

Therefore, 105.33Kg of O2 is needed for the reaction.

5 0
3 years ago
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