Sphalerite contains about 30% zinc. Calculate the maximum amount of zinc that could be obtained from 1000kg sphalerite.

A 57.0 mL sample of a 0.120 M potassium sulfate solution is mixed with 35.5 mL of a 0.118 M lead(II) acetate solution and the fo

Katarina [22]

Answer:

Limiting reagent = lead(II) acetate

Theoretical yield = 1.2704 g

% yield = 78.09 %

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For potassium sulfate :

Molarity = 0.120 M

Volume = 57.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 57.0×10⁻³ L

Thus, moles of potassium sulfate:

$Moles=0.120 M \times {57.0\times 10^{-3}}\ moles$

Moles of potassium sulfate = 0.00684 moles

For lead(II) acetate :

Molarity = 0.118 M

Volume = 35.5 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 35.5×10⁻³ L

Thus, moles of lead(II) acetate :

$Moles=0.118 \times {35.5\times 10^{-3}}\ moles$

Moles of lead(II) acetate = 0.004189 moles

According to the given reaction:

$K_2SO_4_{(aq)}+Pb(C_2H_3O_2)_2_{(aq)}\rightarrow 2KC_2H_3O_2_{(s)}+PbSO_4_{(aq)}$

1 mole of potassium sulfate react with 1 mole of lead(II) acetate

0.00684 moles potassium sulfate react with 0.00684 mole of lead(II) acetate

Moles of lead(II) acetate = 0.004189 moles

Limiting reagent is the one which is present in small amount. Thus, lead(II) acetate is limiting reagent. ( 0.004189 < 0.00684)

The formation of the product is governed by the limiting reagent. So,

1 mole of lead(II) acetate gives 1 mole of lead(II) sulfate

0.004189 mole of lead(II) acetate gives 0.004189 mole of lead(II) sulfate

Molar mass of lead(II) sulfate = 303.26 g/mol

Mass of lead(II) sulfate = Moles × Molar mass = 0.004189 × 303.26 g = 1.2704 g

Theoretical yield = 1.2704 g

Given experimental yield = 0.992 g

% yield = (Experimental yield / Theoretical yield) × 100 = (0.992/1.2704 g) × 100 = 78.09 %

3 0

2 years ago

Compared with the freezing-point depression of a 0.01 m c6h12o6 solution, the freezing-point depression of a 0.01 m nacl solutio

agasfer [191]

Answer:

Twice as much.

Explanation:

That's because the freezing point depression depends on the total number of solute particles.

C₆H₁₂O₆(s) ⟶ C₆H₁₂O₆(aq)

0.01 mol of C₆H₁₂O₆ gives 0.01 mol of solute particles.

NaCl(s) ⟶ Na⁺(aq) + Cl⁻(aq)

1 mol of NaCl gives 0.01 mol of Na⁺(aq) and 0.01 mol of Cl⁻(aq).

That's 0.02 mol of particles, so the freezing point depression of 0.01 mol·L⁻¹ NaCl will be twice that of 0.01 mol·L⁻¹ C₆H₁₂O₆.

6 0

2 years ago

Single strand RNA is made using the genetic instructions from double strand DNA.

ale4655 [162]

False I’m pretty sure

3 0

2 years ago

You are counting red blood cells, which objective will be best to count the number of cells, high power or low power objective a

BARSIC [14]

Answer:

Low power

Explanation:

Low power would allow for the full image of the red blood cells and would appear as small circles.

4 0

2 years ago

A sample of calcium phosphate was found to have a mass of 125.3 g. How many molecules were contained in the sample?

Viktor [21]

The answer for the following problem is mentioned below.

Therefore number of molecules(N) present in the calcium phosphate sample are 19.3 × 10^23 molecules.

Explanation:

Given:

mass of calcium phosphate ( $Ca_{3}(PO_{4} )_{2}$ ) = 125.3 grams

We know;

molar mass of calcium phosphate ( $Ca_{3}(PO_{4} )_{2}$ ) = (40×3) + 3 (31 +(4×16))

molar mass of calcium phosphate ( $Ca_{3}(PO_{4} )_{2}$ ) = 120 + 3(95)

molar mass of calcium phosphate ( $Ca_{3}(PO_{4} )_{2}$ ) = 120 +285 = 405 grams

We also know;

No of molecules at STP conditions( $N_{A}$ ) = 6.023 × 10^23 molecules

To solve:

no of molecules present in the sample(N)

We know;

$\frac{m}{M} =\frac{N} }{}$ N÷ $N_{A}$

$\frac{405}{125.3}$ = $\frac{N}{6.023*10^23}$

N =(405×6.023 × 10^23) ÷ 125.3

N = 19.3 × 10^23 molecules

Therefore number of molecules(N) present in the calcium phosphate sample are 19.3 × 10^23 molecules

3 0

2 years ago