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2 years ago

4. How many grams of hydrogen is produced from 12.5 g of Mg reacting with hydrochloric acid in this

Chemistry

1 answer:

Volgvan2 years ago

8 0

From the reactions, 1.04 g of H2 and 7.995 g of aluminum phosphate is produced.

<h3>What is stoichiometry?</h3>

The term stoichiometry has to do with the amount of substances that participates in a reaction.

For reaction 1;

Mg + 2HCl → MgCl₂ + H₂

Number of moles of Mg reacted = 12.5 g/24g/mol = 0.52 moles

If 1 mole of Mg produced 1 mole of H2

0.52 moles produces 0.52 moles of H2

Mass of H2 = 0.52 moles * 2 g/mol = 1.04 g

For reaction 2;

2Li3PO4 + Al2(SO4)3 → 3Li2SO4 + 2AIPO4

Number of moles of lithium phosphate = 7.5 g/116 g/mol = 0.065 moles

2 moles of Li3PO4 produced 2 moles of AIPO4

0.065 moles of Li3PO4 produced 0.065 moles of AIPO4

Mass of AIPO4 = 0.065 moles * 123 g/mol = 7.995 g

Learn more about stoichiometry:brainly.com/question/9743981

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Question 4

elixir [45]

Answer:

True

Explanation:

4 0

3 years ago

A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and

Romashka [77]

<u>Answer:</u> The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

<u>Explanation:</u>

To calculate the molality of solution, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute $(S_8)$ = 0.800 g

$M_{solute}$ = Molar mass of solute $(S-8)$ = 256.52 g/mol

$W_{solvent}$ = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

$\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m$

<u>Calculation for freezing point of solution:</u>

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

$\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

or,

$\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm$

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC$

Hence, the freezing point of solution is 16.5°C

<u>Calculation for boiling point of solution:</u>

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

or,

$\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm$

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC$

Hence, the boiling point of solution is 118.2°C

8 0

3 years ago

Rolls of foil are 300mm wide and 2.020mm thick. (The density of foil is 2.7 g/cm^3). What maximum length of foil can be made fro

horrorfan [7]

Answer: 8556 mm, or 855.6 cm (8560 mm to 3 sig figs)

Explanation: Convert mm to cm by dividing by 10 (1cm/10mm)

Find the area of the foil face in cm^2 (30cm*0.2020cm) = 0.606 cm^2

Calculate the volume occupied by 1.40 kg of foil in cm^3. 1.40kg = 1400g

1.400g/(2.7 g/cm^3) = 518.5 cm^3 for 1.40 kg Au

Volume = Area (of the face) * Length

We want Length:

Length = Volume/Area

L = (518.5 cm^3/0.606 cm^2)

L = 855.6 cm (8556 mm) Round to 3 sig figs (856 cm and 8560 mm)

5 0

3 years ago

On the basis of the Ksp values below, what is the order of the solubility from least soluble to most soluble for these compounds

Viktor [21]

Answer:

The order of solubility is AgBr < Ag₂CO₃ < AgCl

Explanation:

The solubility constant give us the molar solubilty of ionic compounds. In general for a compound AB the ksp will be given by:

Ksp = (A) (B) where A and B are the molar solubilities = s² (for compounds with 1:1 ratio).

It follows then that the higher the value of Ksp the greater solubilty of the compound if we are comparing compounds with the same ionic ratios:

Comparing AgBr: Ksp = 5.4 x 10⁻¹³ with AgCl: Ksp = 1.8 x 10⁻¹⁰, AgCl will be more soluble.

Comparing Ag2CO3: Ksp = 8.0 x 10⁻¹² with AgCl Ksp = AgCl: Ksp = 1.8 x 10⁻¹⁰ we have the complication of the ratio of ions 2:1 in Ag2CO3, so the answer is not obvious. But since we know that

Ag2CO3 ⇄ 2 Ag⁺ + CO₃²₋

Ksp Ag2CO3 = 2s x s = 2 s² = 8.0 x 10-12

s = 4 x 10⁻12 ∴ s= 2 x 10⁻⁶

And for AgCl

AgCl ⇄ Ag⁺ + Cl⁻

Ksp = s² = 1.8 x 10⁻¹⁰ ∴ s = √ 1.8 x 10⁻¹⁰ = 1.3 x 10⁻⁵

Therefore, AgCl is more soluble than Ag₂CO₃

The order of solubility is AgBr < Ag₂CO₃ < AgCl

8 0

3 years ago

Read 2 more answers

The one i selected is wrong!! pls help:(( only answer if ur right

Darina [25.2K]

Answer:

Lithium's mass number would decrease by 1

Explanation:

7 0

3 years ago