It increases with. And

If our aim is to make 100% pure alum, how could large chunks be bad?

iren [92.7K]

Answer:

It prevent the solution from making 100% pure alum.

Explanation:

Large chunks could be bad if we make 100% pure alum because the large chunks prevent the production of 100% pure alum so for making 100% pure alum, large chunks will be removed from the solution or process. Alum is a chemical compound having salt of sulfate attached with aluminum in hydrated form i. e. presence of water. It is widely used in vaccines and for the purification of drinking water.

8 0

3 years ago

If you have 3.54 g of H2, how many grams of NH3 can be produced?

morpeh [17]

3.54g H2 means you have 3.54 / 2 = 1.77 moles of H2, which means you have 1.77*2 = 3.54 moles of H atom.

Since 1 mole of NH3 has 3 moles of H, so you can produce 3.54 / 3 moles of NH3, which is (3.54/3) * (12+3) = 17.7 grams

5 0

3 years ago

. The Henry's law constant for helium gas in water at 30 ∘C is 3.7×10−4M/atm; the constant for N2 at 30 ∘C is 6.0×10−4M/atm. a.

ludmilkaskok [199]

Explanation:

1) Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$M_{g}=K_H\times p_{g}$

where,

$K_H$ = Henry's constant

$p_{He}$ = partial pressure of gas

a) $K_H=3.7\times 10^{-4} M/atm$

$p_{He}=2.1 atm$

Putting values in above equation, we get:

$M_{He}=3.7\times 10^{-4} M/atm\times 2.1 atm = 7.77\times 10^{-4} M$

The solubility of helium gas is $7.77\times 10^{-4} M$

b) $K_H=6.0\times 10^{-4} M/atm$

$p_{N_2}=2.1 atm$

Putting values in above equation, we get:

$M_{He}=6.0\times 10^{-4} M/atm\times 2.1 atm = 1.26\times 10^{-3} M$

The solubility of nitrogen gas is $1.26\times 10^{-3} M$

2)

a) Mass of solute or methanol , m= 14.7 g

Mass of solvent or water , m'= 186 g

Mass of the solution = M = m + m' = 14.7 g + 186 g = 200.7 g

$w/w\%=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100$

$=\frac{14.7 g}{200.7 g}\times 100=7.32\%$

The mass percent of methanol is 7.32%.

b) Molality is defined as moles of solute per kilograms of solvent.

$m=\frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}$

Moles of methanol = $\frac{14.7 g}{32 g/mol}=0.4594 mol$

Mass of solvent = 186 g = 0.186 kg

$m=\frac{0.4594 mol}{0.186 kg}=2.4610 mol/kg$

The molality of methanol is 2.4610 mol/kg.

4 0

3 years ago

(50 POINTS)THIS IS A TIMED TEST PLEASE ANSWER

aliina [53]

Answer:

the second one down is your answer

Explanation:

5 0

3 years ago

An aqueous barium chloride solution was added to a 15.0 mL sample of a sodium sulfate solution according to the reaction below.

Dimas [21]

<h3>Answer:</h3>

0.523 M

<h3>Explanation:</h3>

The reaction between barium chloride and sodium sulfate is given by

BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq)

We are given,

Volume of Na₂SO₄ = 15.0 mL

Mass of of solid BaSO₄ = 1.83 g

Required to determine the molarity of Na₂SO₄ solution

we will use the following steps

<h3>Step 1: Determine moles of the solid BaSO₄</h3>

Mass of BaSO₄ = 1.83 g

To get the number of moles we divide mass by the molar mass

Molar mass of BaSO₄ = 233.38 g/mol

Number of moles = 1.83 g ÷ 233.38 g/mol

= 0.00784 moles

<h3>Step 2: Moles of sodium sulfate used </h3>

From the balanced equation for every 1 mole of sodium sulfate used 1 mole of BaSO₄ was produced.

Therefore, the mole ratio of Na₂SO₄ : BaSO₄ is 1 : 1

Hence, moles of Na₂SO₄ will also be 0.00784 moles

<h3>Step 3: Molarity of sodium sulfate solution </h3>

Volume of sodium sulfate = 15.0 mL or 0.015 L

Number of moles = 0.00784 moles

But, molarity = Moles ÷ Volume

= 0.00784 moles ÷ 0.015 L

= 0.5227 M

= 0.523 M

Thus, the molarity of original sodium sulfate is 0.523 M

7 0

3 years ago