All categories

1 year ago

Determine the mass of oxygen in a 7.20 g sample of Al2(SO4)3.

Chemistry

2 answers:

Mekhanik [1.2K]1 year ago

5 0

Given:

7.20 g sample of Al2(SO4)3

Required:

Mass of oxygen

Solution:

Since you are not given a chemical reaction, just base your solution to the chemical formula given.

Molar mass of Al2(SO4)3 = 342.15 g/mol

7.20 g Al2(SO4)3 (1 mol/342.15g)(3mol O/2 mol Al)(1 mol O2/1/2 mol O2)(32g O2/1mol O2) = 4.04 g O2

Anarel [89]1 year ago

3 0

7.20 g of a sample of Al2(SO4)3 contains 4.04 g of oxygen.

<h2>Further Explanation</h2><h3>A compound </h3>

A compound is a substances that contains two or more different atoms that are bonded together. When the atoms are similar the substance is known as a molecule, therefore not all molecules are compounds.

<h3>Molar mass </h3>

Molar mass is the mass of one mole of a compound or a substance. The molar mass of a compound is the mass that is equivalent to the sum of relative atomic masses of the atoms of elements making up the compound.

<h3>Mass percent composition of a compound </h3>

Mass percent composition of a compound is calculated by dividing the mass of element in one mole of the compound by the molar mass of the compound and multiplying by 100%.

In this case we are require to calculate the mass of Oxygen in 7.2 g of Al2(SO4)3

Therefore; the molar mass of Al2(SO4)3 will be; 342.15 g/mol
Al2(SO4)3 is made up of two Al atoms, three sulfur atoms, and 12 oxygen atoms.

Therefore;

1 mole of Al2(SO4)3 contains a mass of 192 g of Oxygen

Thus; A sample of 7.20 g Al2(SO4)3 will contain;

= ((7.2 g Al2(SO4)3x(192 g O))/ 342.15 g/mol Al2(SO4)3

= 4.04 g Oxygen

Keywords: Molar mass, compound

<h3>Learn more about: </h3>

Compounds: brainly.com/question/1194814
Atoms: brainly.com/question/12505687
Percentage composition by mass in compounds: brainly.com/question/9118852

Level: High school

Subject: Chemistry

Topic: Compounds and mixtures

Sub-topic: Composition of compounds

You might be interested in

Calculate the enthalpy of combustion, δh∘comb, for c6h14. you'll first need to determine the balanced chemical equation for the

Vedmedyk [2.9K]

<span>Answer: Enthalpy Change = (6 x -393.5) + (7 x -285.8) - (-204.6) + (19/2) 0.....??? like.. (6 x Enth CO2) + ( 7 x Enth H2O) - (Enth C6H14) + (19/2) Enth O2</span>

3 0

2 years ago

How many moles of gas Does it take to occupy 520 mL at a pressure of 400 torr and a temperature of 340 k

Ann [662]

Answer would be B. I provided work on an image attached. Message me if u have any other questions on how to do it

6 0

1 year ago

You prepare a standard by weighing 10.751 mg of compound X into a 100 mL volumetric flask and making to volume. You further dilu

ArbitrLikvidat [17]

Answer:

0.12693 mg/L

Explanation:

First we <u>calculate the concentration of compound X in the standard prior to dilution</u>:

10.751 mg / 100 mL = 0.10751 mg/mL

Then we <u>calculate the concentration of compound X in the standard after dilution</u>:

0.10751 mg/mL * 5 mL / 25 mL = 0.021502 mg/L

Now we calculate the<u> concentration of compound X in the sample</u>, using the <em>known concentration of standard and the given areas</em>:

2582 * 0.021502 mg/L ÷ 4374 = 0.012693 mg/L

Finally we <u>calculate the concentration of X in the sample prior to dilution</u>:

0.012693 mg/L * 50 mL / 5 mL = 0.12693 mg/L

4 0

1 year ago

If an equal number of moles of reactants are used, do the following equilibrium mixtures contain primarily reactants or products

puteri [66]

<u>Answer:</u>

<u>For a:</u> The equilibrium mixture contains primarily reactants.

<u>For b:</u> The equilibrium mixture contains primarily products.

<u>Explanation:</u>

There are 3 conditions:

When $K_{eq}>1$ ; the reaction is product favored.
When $K_{eq}$ ; the reaction is reactant favored.
When $K_{e}=1$ ; the reaction is in equilibrium.

For the given chemical reactions:

<u>For a:</u>

The chemical equation follows:

$HCN(aq.)+H_2O(l)\rightleftharpoons CN^-(aq.)+H_3O^+(aq.);K_{eq}=6.2\times 10^{-10}$

The expression of $K_{eq}$ for above reaction follows:

$K_{eq}=\frac{[CN^-][H_3O^+]}{[HCN][H_2O]}=6.2\times 10^{-10}$

As, $K_{eq}$ , the reaction will be favored on the reactant side.

Hence, the equilibrium mixture contains primarily reactants.

<u>For b:</u>

The chemical equation follows:

$H_2(g)+Cl_2(g)\rightleftharpoons 2HCl(g);K_{eq}=2.51\times 10^{4}$

The expression of $K_{eq}$ for above reaction follows:

$K_{eq}=\frac{[HCl]^2}{[H_2][Cl_2]}=2.51\times 10^{4}$

As, $K_{eq}>1$ , the reaction will be favored on the product side.

Hence, the equilibrium mixture contains primarily products.

4 0

1 year ago

A vessel of capacity 400 cc filled with chlorine under 80 cm is connected by a narrow tube and stopcock with another vessel of c

guapka [62]

Answer:

The pressure when the stopcock is opened is opened is 87.783 cm

Explanation:

The given parameters of the question are;

The volume of the vessel of chlorine = 400 cc

The pressure of the vessel of chlorine = 80 cm

The volume of the vessel of chlorine = 250 cc

The pressure of the vessel of chlorine = 100 cm

Daltons law of Partial Pressures states that the total pressure exerted by a volume of a mixture of gases is equal to the partial pressures exerted by the individual gases in the mixture with respect to the given volume

Therefore;

The total volume of the mixture = 400 cc + 250 cc = 650 cc

The partial pressure exerted by the chlorine gas in the total volume is given by Boyles law as follows;

P₁·V₁ = P₂·V₂

P₂ = P₁·V₁/V₂

Where;

P₁ = 80 cm = The pressure in volume V₁ = 400 cc

P₂ₓ = The partial pressure of chlorine in volume V₂ = 650 cc

Substituting, we have;

P₂ₓ = 80 × 400/650 ≈ 49.321 cm

Similarly, the partial pressure exerted by the nitrogen gas in the total volume is given by Boyles Law as follows;

P₂ₐ = P₁·V₁/V₂

Where;

P₁ = 100 cm = The pressure in volume V₁ = 250 cc

P₂ₐ = The partial pressure of nitrogen in volume V₂ = 650 cc

Substituting, we have;

P₂ₐ = 100× 250/650 ≈ 38.462 cm

The pressure of the combined gas, P, when the stopcock is opened is opened is given by Dalstons Law of partial pressure as P = P₂ₐ + P₂ₓ

Therefore, the pressure, P when the stopcock is opened is opened = 49.321 cm + 38.462 cm = 87.783 cm

3 0

1 year ago