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2 years ago

Find the percent composition of OXYGEN in Manganese (III) nitrate, Mn(NO3)3.

Chemistry

1 answer:

BaLLatris [955]2 years ago

8 0

Answer:

59.8%

Explanation:

First find the Mr of manganese (III) nitrate.

Mr of Mn(NO₃)₃ = 54.9 + (14 × 3) + (16 × 3 × 3) = 240.9

Since we have to find the percentage composition of oxygen, we need to find the Mr of oxygen in the compound, which is:

Mr of (O₃)₃ = (16 × 3) × 3 = 144

Now we can find percentage composition / percentage by mass of oxygen.

% composition = $\frac{Mr\ of\ oxygen\ in\ compound}{Mr\ of\ compound}$ × 100

% composition = $\frac{144}{240.9}$ × 100 = 59.776%

∴ % compostion of oxygen in maganese(III)nitrate is 59.8% (to 3 significant figures).

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Dmitrij [34]

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3 years ago

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O tetraclorometano é um solvente muito utilizado na indústria de transformaç?o de produtos orgânicos ele é produzido pela reaç?o

Vlada [557]

Answer:

ΔHr = -103,4 kcal/mol

Explanation:

Using:

AH° (kcal/mol)

Metano (CH)

-17,9

Cloro (CI)

tetraclorometano (CCI)

- 33,3

Acido cloridrico (HCI)

-22

It is possible to obtain the ΔH of a reaction from ΔH's of formation for each compound, thus:

ΔHr = (ΔH products - ΔH reactants)

For the reaction:

CH₄(g) + Cl₂(g) → CCl₄(g) + HCl(g)

The balanced reaction is:

CH₄(g) + 4Cl₂(g) → CCl₄(g) + 4HCl(g)

The ΔH's of formation for these compounds are:

ΔH CH₄(g): -17,9 kcal/mol

ΔH Cl₂(g): 0 kcal/mol

ΔH CCl₄(g): -33,3 kcal/mol

ΔH HCl(g): -22 kcal/mol

The ΔHr is:

-33,3 kcal/mol × 1 mol + -22 kcal/mol× 4 mol - (-17,9 kcal/mol × 1 mol + 0kcal/mol × 4mol)

ΔHr = -103,4 kcal/mol

I hope it helps!

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3 years ago

If a flask containing one liter of water was poured into 1,000 small cubes, then each cube would contain:

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Every cube would contain 1 milliliter of water, or .01 liter

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3 years ago

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Select all of the following statements that are false about ΔGo and ΔG:a) If the reaction has a large negative ΔGo value, the re

OLEGan [10]

Answer:

a) If the reaction has a large negative ΔGo value, the reaction must reach equilibrium at a small extent of reaction value

d) ΔGo and ΔG have the same magnitude, they just have opposite signs.

Explanation:

The fraction of the total heat energy if a system that does useful work is known as Gibb's free energy (G) and the change from the initial to final state is designated by $\Delta G$ . It is observed that the values of $\Delta G$ changes with experimental conditions such as temperature , pressure , concentration etc.

$\Delta G^0$ is the standard free energy change which is a balance of two natural tendencies of any system.

Minimization of potential energy or enthalpic factor $\Delta H^0$

Maximization of disorderliness or entropic factor $T\Delta S^0$

Mathematically; $\Delta G$ = $\Delta H^0$ - $T\Delta S^0$

Thus; from above mentioned, the statements that are true about ΔG⁰ and ΔG are:

ΔG⁰ and ΔG can have different values, they don't even have to have the same sign

For a reaction that reaches equilibrium, the minimum value of free energy must be at the equilibrium point

If ΔG⁰ , measured at an extent of reaction = 0.5, is positive, the sign for ΔG when the extent of reaction = 0.80 is also positive.

while the false statements include:

a) If the reaction has a large negative ΔG⁰ value, the reaction must reach equilibrium at a small extent of reaction value

d) ΔG⁰ and ΔG have the same magnitude, they just have opposite signs.

7 0

3 years ago

A 2.350×10−2 M solution of NaCl in water is at 20.0∘C. The sample was created by dissolving a sample of NaCl in water and then b

sveta [45]

Answer:

Part A: m = 0.02356 mol/kg = 0.02356 m

Part B: Xsolute = 4.243×10⁻⁴

Part C: % m/m = 0.1376%

Part D: ppm = 1,376 ppm

Explanation:

1. Data:

a) M = 2.350×10⁻² M

b) V sol = 1.000 L

c) V H₂O = 994.4 mL = 0.9944 L

d) d H₂O = 0.9982 g/mL

2. Formulae

M = n solute / V sol (L)
m = n solute / Kg solvent
X solute = n solute / N total
% m/m = (mass of solute / mass of solution) × 100
ppm = (mass of solute / mass of solution) × 1,000,000
density = mass in grams / volume in mL

3. Solution

Part A: Calculate the molality of the salt solution.

m = n solute / Kg solvent

i) M = n solute / V sol (L) ⇒ n solute = M × V sol (L)

⇒ n solute = M = 2.350×10⁻² M × 1.000 L = M = 2.350×10⁻² mol

ii) density H₂O = mass H₂O / volume H₂O

⇒ mass H₂O = density H₂O × volume H₂O

⇒ mass H₂O = 0.9982 g/mL × 999.4 mL = 997.6 g

iii) kg H₂O = 997.6 g / (1,000 g/Kg) = 0.9976 kg

iv) m = 2.350×10⁻² mol / 0.9976 kg = 0.02356 mol/kg = 0.02356 m

Part B: Calculate the mole fraction of salt in this solution.

X solute = n solute / N total

i) n solute = 2.350×10⁻² mol

ii) n solvent = n H₂O = mass H₂O in grams/ molar mass H₂O

⇒ 997.6 g / 18.015 g/mol = 55.38 mol

iii) X solute = 2.350×10⁻² mol / 55.38 mol = 4.243×10⁻⁴

Part C: Calculate the concentration of the salt solution in percent by mass.

% m/m = (mass of solute / mass of solution) × 100

i) molar mass = mass in grams / molar mass

⇒ mass of solute = mass of NaCl = n solute × molar mass NaCl

⇒ mass of solute = 2.350×10⁻² mol × 58.44 g/mol = 1.373 g

ii) % m/m = (1.373 g / 997.6 g) × 100 = 0.1376%

Part D: Calculate the concentration of the salt solution in parts per million.

ppm = (mass of solute / mass of solution) × 1,000,000

i) ppm = ( (1.373 g / 997.6 g) × 1,000,000 = 1,376 ppm

5 0

3 years ago