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nasty-shy [4]
3 years ago
14

How many milliliter of a solution of 4.00KI are needed to prepare 250.0mL of 0.760 KI

Chemistry
1 answer:
Alexeev081 [22]3 years ago
3 0
Answer:

47.5 mL

Solving:

M1 = 4.00 M

V1 = ?

M2 = 0.760 M

V2 = 0.250 L

---

M1 * V1 = M2 * V2

V1 = ( M2 * V2 ) / M1

V1 = ( 0.760 * 0.250 ) / 4.00

V1 = ( 0.190 ) / 4.00

V1 = 0.0475 L
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A year ago If an element has 12 protons and 17 neutrons, how many electrons must it have?
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There are three naturally occurring isotopes of the hypothetical element hilarium 45Hi, 46Hi, and 48Hi. The percentages of these
den301095 [7]

Answer:

46.761g/mol

Explanation:

Given parameters:

Element = Hilarium , Hi

Isotopes: Hi- 45, Hi-46 and Hi- 48

Natural abundance of Hi-45 = 18.3%

                                     Hi-46 = 34.5%

                                     Hi-48 = 47.2%

Unknown:

Atomic weight of naturally occurring Hilarium = ?

Solution:

Isotopes have been studied extensively by mass spectrometry. The method is used to determine the proportion/percentage/fraction by which each of the isotopes of an element occurs in nature. The proportion is called geonormal abundance. From this we can calculate the atomic weight of an element.

 We can use the expression below to find this value:

       Atomic weight = m₄₅α₄₅ + m₄₆α₄₆ + m₄₈α₄₈

    m is the atomic mass of each isotope and α is the abundance

Atomic weight = (45 x \frac{18.3}{100} ) + (46 x  \frac{34.5}{100} ) + (48 x  \frac{47.2}{100})

Atomic weight of Hi = 8.235 + 15.870 +  22.656 = 46.761g/mol

6 0
3 years ago
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
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