Volume is the independent or dependent variable

Which formula can be used to calculate the molar mass of hydrogen peroxide (H2O2)? (5 points)

ratelena [41]

answer: option d (2×molar mass of H +2×molar mass of O

 

7 0

2 years ago

What is the value of delta Hrxn for this equation:

Ratling [72]

Answer:

The ΔHrxn for the above equation = 179 kJ/mol

Explanation:

The reaction bond enthalpies are for the reactant;

3 × N-H = 3 × 390 = 1,170 kJ/mol

2 × O=O = 2 × 502 = 1004 kJ/mol

The reaction bond enthalpies are for the product;

3 × N-O = 3 × 201 = 603 kJ/mol

3 × O-H = 3 × 464 = 1,392 kJ/mol

The ΔHrxn for the above equation is therefore;

ΔHrxn = 1,170 + 1,004 - (603 + 1,392) = 179 kJ/mol

6 0

2 years ago

A compound is composed of C, H and O. A 1.621 g sample of this compound was combusted, producing 1.902 g of water and 3.095 g of

vlada-n [284]

Answer: The molecular of the compound is, $C_2H_3O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=3.095g$

Mass of $H_2O=1.902g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in $3.095g$ of carbon dioxide, $\frac{12}{44}\times 3.095=0.844g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in $1.902g$ of water, $\frac{2}{18}\times 1.092=0.121g$ of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = $(1.621)-[(0.844)+(0.121)]=0.656g$

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.844g}{12g/mole}=0.0703moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.121g}{1g/mole}=0.121moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.656g}{16g/mole}=0.041moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $0.041$ moles.

For Carbon = $\frac{0.0703}{0.041}=1.71\approx 2$

For Hydrogen = $\frac{0.121}{0.041}=2.95\approx 3$

For Oxygen = $\frac{0.041}{0.041}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 3 : 1

Hence, the empirical formula for the given compound is $C_2H_3O_1=C_2H_3O$

The empirical formula weight = 2(12) + 3(1) + 1(16) = 43 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{46.06}{43}=1$

Molecular formula = $(C_2H_3O_1)_n=(C_2H_3O_1)_1=C_2H_3O$

Therefore, the molecular of the compound is, $C_2H_3O$

6 0

2 years ago

A sample of gas in which [h2s] = 3.50 m is heated to 1400 k in a sealed vessel. after chemical equilibrium has been achieved, wh

Bond [772]

The value of Kc for the thermal decomposition of H₂S is 2.2 x 10⁻⁴ at 1400 K:
2 H₂S(g) ↔ 2 H₂(g) + S₂(g)
initial 3.5 M 0 0
at equilibrium 3.5 M - 2x 2x x
Kc = [S₂][H₂]² / [H₂S]²
2.2 X 10⁻⁴ = x(2x)² / (3.5 - 2x)²
2.2 x 10⁻⁴ = 4 x³ / (3.5)² Assuming x <<<<< 3.5
x = 0.088
Thus [H₂S] = 3.324 M

6 0

3 years ago

600.0 mL of air is at 20.0 c what is the volume at 60.0

pogonyaev

682mL

Explanation:

Given parameters:

Initial volume of air = 600mL

Initial temperature = 20°C

Final temperature = 60°C

Unknown:

Final volume = ?

Solution:

To solve this problem, we apply Charles's law';

Charles's law states that "at constant pressure, the volume of a given mass of gas is directly proportional to its temperature. "

Mathematically;

$\frac{V_{1} }{T_{1} } = \frac{V_{2} }{T_{2} }$

V₁ is the initial volume of air

T₁ is the initial temperature of air

V₂ is the final volume of air

T₂ is the final temperature of air

To proceed in solving this problem, we need to convert the given temperature to Kelvin;

T K = 273 + T°C

T₁ = 273 + 20 = 293K

T₂ = 273 + 60 = 333K

now input the parameters;

$V_{2} = \frac{V_{1} }{T_{1} } x T_{2} = \frac{600}{293 } x 333$

V₂ = 682mL

learn more:

Gas laws brainly.com/question/2438000

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8 0

3 years ago