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Harlamova29_29 [7]
2 years ago
7

Which sample of matter is a mixture (1) Br2(l) (2) K(s) (3) KBr(s) (4)KBr(aq)

Chemistry
1 answer:
SOVA2 [1]2 years ago
5 0

Answer:

(4)KBr(aq)

Explanation:

A mixture is an impure substance made up of two or more substances that are joined together. Their composition is indefinite and not easily represented by a simple chemical formula.

KBr_{aq} is a mixture because it made up of an aqueous solution of KBr. This suggests that the substance KBr is placed inside water.

Here, we have two states of matter which is the solid and the liquid coming together.

As with both mixtures, they can be separated by physical means, KBr aqueous solution can also be done this way.

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Pls help!! will give brainly
____ [38]

Answer:

True

Explanation:

4 0
3 years ago
Read 2 more answers
The generic metal hydroxide M(OH)2 has Ksp = 8.05×10−12. (NOTE: In this particular problem, because of the magnitude of the Ksp
vodka [1.7K]

Answer:

a. in pure water Solubility (x) = 1.26 x 10⁻⁴M

b. in 0.202M M⁺² Solubility (x) = 9.963 x 10⁻¹²M

The large drop in solubility is consistent with the common ion effect.

Explanation:

a. Solubility in pure water

Given:       M(OH)₂ ⇄ M⁺² + 2OH⁻

            I       ---            0          0

           C      ---             x          2x

           E       ---            x           2x

Ksp = [M⁺²][OH⁻]² = (x)(2x)² = 4x³ => x = CubeRt(Ksp/4)

solubility in pure water = x = CubeRt(8.05 x 10⁻¹²/4) = 1.26 x 10⁻⁴M

b. Solubility in presence of 0.202M M⁺² as common ion.

Given:       M(OH)₂ ⇄       M⁺²      +     2OH⁻

            I       ---            0.202M              0

           C      ---                 +x                 +2x

           E       ---           0.202M + x         2x

                                   ≈ 0.202M

Ksp = [M⁺²][2x]² = (0.202)(2x)² = (0.202)(4x²) = 8.05 x 10⁻¹²

=> x = (8.05 x 10⁻¹²)/(0.202)(4) = 9.963 x 10⁻¹²M

6 0
2 years ago
A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0 mL of 1 M H2SO4. A 25.00 mL aliquot is anal
Olenka [21]

Answer:

The weight percent in the sample is 17,16%

Explanation:

The dissolution of the Ce(IV) salt provides free Ce⁴⁺ that reacts, thus:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃²⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03247M Na₂S₂O₃ = 4,228x10⁻⁴ moles of S₂O₃²⁻.

As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,228x10⁻⁴ moles of S₂O₃⁻×\frac{1molI_{3}^-}{2molS_{2}O_{3}^{2-}} = <em>2,114x10⁻⁴ moles of I₃⁻</em>

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,114x10⁻⁴ moles of I₃⁻× \frac{2molCe^{4+}}{1molI_{3}^-} =  <em>4,228x10⁻⁴ moles of Ce(IV)</em>.

These moles are:

4,228x10⁻⁴ moles of Ce(IV)×\frac{140,116g}{1mol} = 0,05924 g of Ce(IV)

As was taken an aliquot of 25,00mL from the solution of 250,0mL:

0,05924 g of Ce(IV)×\frac{250,0mL}{25,00mL} =0,5924g of Ce(IV) in the sample

As the sample has 3,452g, the weight percent is:

0,5924g of Ce(IV) / 3,452g × 100 = <em>17,16 wt%</em>

I hope it helps!

3 0
3 years ago
find the volume of an item using a scale and waterA crown made of alloy of gold and silver has a volume of 60cm3and mass of 1050
KiRa [710]

Answer:

The mass of gold in crown is 960.61 g.And  the volume of gold is 49.77 cm^3.

Explanation:

Density is defined as mass of the substance present in unit volume of the substance.

Mass of gold in an alloy of crown= m

Mass of silver in an alloy of crown = M

Volume  gold in an alloy of crown= v

Volume of silver in an alloy of crown = V

Density of gold = 19.3 g/cm^3

Density of silver = 10.5 g/cm^3

Volume of the crown = 60 cm^3

60 cm^3=v+V

60 cm^3=\frac{m}{19.3 g/cm^3}+\frac{M}{10.5 g/cm^3}..(1)

Mass of eh crown = 1050 g

1050g =m+M..(2)

Solving equation (1) and (2):

m = 960.61 g

M = 89.39 g

Volume of the gold = v = \frac{960.61 g}{19.3 g/cm^3}=49.77 cm^3

Volume of silver = V = 60 cm^3-v=60 cm63-49.77 cm63=10.23 cm^3

The mass of gold in crown is 960.61 g.And  the volume of gold is 49.77 cm^3.

4 0
3 years ago
How much heat is absorbed by a 2000 g granite rock as energy from the sun causes its temperature to change from 10°C to 29°C?
emmainna [20.7K]

Answer:

None of the options are correct. The correct answer is 30020J

Explanation:

Data obtained from the question include:

Mass (M) = 2000g

Initial temperature (T1) = 10°C

Final temperature (T2) = 29°C

Change in temperature (ΔT) = T2 - T1 = 29°C - 10°C = 19°C

Heat (Q) =?

Specific heat capacity (C) of granite = 0.79J/g°C

Applying the equation Q = MCΔT, the heat absorbed by the granite rock can be obtained as follow:

Q = MCΔT

Q = 2000 x 0.79 x 19

Q = 30020J

3 0
2 years ago
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