All categories

2 years ago

Which of the following compounds are soluble in aqueous solutions? Select ALL that are soluble.

Chemistry

1 answer:

sattari [20]2 years ago

7 0

<h3>Answer:</h3>

K₂S
CaSO₄ (slightly soluble in water)
Pb(ClO₄)₂

<h3>Explanation:</h3>

Mg(OH)₂ and CoCO₃ are insoluble in water.
Solubility rules are used to determine whether a compound will be soluble or not soluble in aqueous form.
K₂S is soluble in aqueous solutions as all salts of Potassium are soluble in water.

Calcium (ii) sulfate is slightly soluble in water.
Hydroxides are insoluble in water except, KOH, NaOH and NH₄OH.and Barium hydroxide. Therefore, Mg(OH)₂ is an insoluble hydroxide.

CoCO₃ is insoluble since carbonates are insoluble except K₂CO₃, Na₂CO₃ and (NH₄)₂CO₃. .

Pb(ClO₄)₂ is soluble because all perchlorates are soluble.

You might be interested in

9. Using the balanced equation from Question #8, how many grams of lead will be produced if 2.54 grams of PbS is burned with 1.8

MissTica

Answer: 2.24 grams of Pb

Explanation:

Balanced chemical reaction;

2PbS + 3O2 → 2Pb + 2SO3

Moles of both PbS and O2

Moles = mass / molar mass

Moles of PbS = 2.54 g / 239.3 g/mol = 0.0108 moles

Moles of O2 = 1.88 / 32 g/mol = 0.0588 moles

Finding the limiting reactant.

Limiting reactant, is that reactant which is completely used in the reaction;

If we assume that PbS is the limiting reactant;

We have 0.0588 moles of O2. This needs ( 0.0588 * 2) / 3 = 0.0392 moles of PbS to fully react. But we have only 0.0108 moles of PbS available. That means that the PbS will be completely consumed hence the limiting reactant

If we assume O2 is the limiting reactant;

We have 0.0108 moles of PbS. That needs ( 0.0108 * 3) / 2 = 0.0162 moles of O2. But we have 0.0588 moles of O2 which is in excess further confirming that PbS is the limiting reactant since it will be depleted in the reaction.

Moles of lead

For this step we apply the mole ratios with the limiting reactant;

Mole ratio of PbS : Pb = 2 : 2 = 1 : 1

Therefore;

Moles of Pb = (0.0108 moles * 1 ) 1

Moles of Pb =0.0108 moles

Mass of Pb

Mass = moles * molar mass

Mass of Pb =0.0108 moles * 207.2 g/mol

Mass of Pb = 2.24 grams

5 0

2 years ago

A container of gas has a volume of 140.0 cm3 at a temperature of 27.0°C. If the

Vitek1552 [10]

Answer:

285.185 (.185 repeating) cm^3

Explanation:

To get the answer, you divide 140 by 27 to get 5.185 (.185 repeating). Then, you multiply 5.185 (.185 repeating) by 55 and get 285.185 (.185 repeating) cm^3. Please use ^ next time to indicate exponents.

8 0

2 years ago

3 Cu + 8HNO3 g 3 Cu(NO3)2 + 2 NO + 4 H2O

liubo4ka [24]

24.25 moles of NO can be produced using 97 moles of HNO3.

<h3>What is balanced chemical equation?</h3>

Equal numbers of atoms from various elements are present in both the reactants and the products in balanced chemical equations. Varied elements' atom counts in the reactants and products of unbalanced chemical equations are different.

3 Cu + 8HNO3 g → 3 Cu(NO3)2 + 2 NO + 4 H2O

The number of moles consumed can be calculated using comparing with coefficients in the balanced reaction .

So , from above eq we get that 8 moles of HNO3 are consumed to make 2 moles of NO.

⇒ 8 HNO3⇔2 NO

⇒ 1 HNO3⇔ 1/4 NO

This means that for each mole of HNO3 produces 1/4 moles of NO.

So , for 97 moles of HNO3 , $\frac{1}{4} *97$ moles of NO can be made,

So, total moles of NO made are 24.25 moles.

Lean more about balanced reactions here brainly.com/question/26694427

#SPJ10

3 0

1 year ago

Read 2 more answers

Help please!<br> 1. L <br> 2. M<br> 3. X<br> 4. Z

Mkey [24]

1. L
Number one because the lines match up

5 0

2 years ago

Express the following in liters at STP:<br><br><br><br> 6.62 x 10-3 moles HF

serious [3.7K]

Answer:

The volume of 6.62×10⁻³moles of HF at STP is 148.38×10⁻³ L

Explanation:

Given data:

Number of moles of HF = 6.62×10⁻³ mol

Volume of HF in litter at STP = ?

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K

T = temperature in kelvin

Standard temperature = 273 K

Standard pressure = 1 atm

Now we will put the values in formula.

1 atm × V = 6.62×10⁻³mol ×0.0821 atm.L/ mol.K × 273 K

V = 6.62×10⁻³mol ×0.0821 atm.L/ mol.K × 273 K / 1 atm

V = 148.38×10⁻³ L

Thus, the volume of 6.62×10⁻³moles of HF at STP is 148.38×10⁻³ L.

8 0

3 years ago