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3 years ago

A 36.41-g sample of calcium carbonate (CaC O 3 )

1 answer:

7 0

M(O) = m(CaCO₃) - m(Ca) - m(C).
m(O) = 36.41 g - 14.58 g - 4.36 g.
m(O) = 17.47 g.
ω(Ca) = m(Ca) ÷ m(CaCO₃) · 100%.
ω(Ca) = 14.58 g ÷ 36.41 g · 100%.
ω(Ca) = 40 %; mass percent of calcium.
ω(C) = m(C) ÷ m(CaCO₃) · 100%.
ω(C) = 4.36 g ÷ 36.41 g · 100%
ω(C) = 12%; mass percent of carbon.
ω(O) = 100% - ω(Ca) - ω(C).
ω(O) = 100% - 40% - 12%.
ω(O) = 48%; mass percent of oxygen.

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F a sample of butene (C4H8) that has a mass of 136.6 g is combusted in excess oxygen, what is the mass of CO2 that is produced?

alukav5142 [94]

The balanced chemical reaction will be:

C4H8 + 6 O2 --> 4 CO2 + 4 H2O

We are given the amount of butene being combusted. This will be our starting point.

136.6 g C4H8 (1 mol C4H8/ 56.11 g C4H8) (4 mol CO2/1 mol C4H8) ( 44.01 g CO2/ 1 mol CO2) = 428.6 g CO2

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3 years ago

Read 2 more answers

Cars run on gasoline, where octane (C8H18) is the principle component. This combustion reaction is responsible for generating en

Bezzdna [24]

Answer:

10.19 g CO₂

4.69 g H₂O

Explanation:

The combustion reaction of Octane is:

C₈H₁₈ → 8CO₂ + 9H₂O

To calculate the mass of CO₂ and H₂O produced, we need to know the mass of octane combusted.

We calculate the mass of Octane from the given volume and density, using the following conversion factors:

1 gallon = 3.785 L

1 L = 1000 mL

Now we convert 1.24 gallons to mL:

1.24 gallon * $\frac{3.785L}{1gallon} *\frac{1000mL}{1L} =$ 4693.4 mL

We calculate the mass of Octane:

4693.4 mL * 0.703 g/mL = 3.30 g Octane

Now we use the stoichiometric ratios and molecular weights to calculate the mass of CO₂ and H₂O:

CO₂ ⇒ 3.30 g Octane ÷ 114g/mol * $\frac{8molCO_{2}}{1molOctane}$ * 44 g/mol = 10.19 g CO₂

H₂O ⇒ 3.30 g Octane ÷ 114g/mol * $\frac{9molH_{2}O}{1molOctane}$ * 18 g/mol = 4.69 g H₂O

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3 years ago

A sample of ammonia ^NH3h gas is completely decomposed to nitrogen and hydrogen gases over heated iron wool. If the total pressu

icang [17]

Answer : The partial pressure of $N_2$ and $H_2$ is, 216.5 mmHg and 649.5 mmHg

Explanation :

According to the Dalton's Law, the partial pressure exerted by component 'i' in a gas mixture is equal to the product of the mole fraction of the component and the total pressure.

Formula used :

$p_i=X_i\times p_T$

$X_i=\frac{n_i}{n_T}$

So,

$p_i=\frac{n_i}{n_T}\times p_T$

where,

$p_i$ = partial pressure of gas

$X_i$ = mole fraction of gas

$p_T$ = total pressure of gas

$n_i$ = moles of gas

$n_T$ = total moles of gas

The balanced decomposition of ammonia reaction will be:

$2NH_3\rightarrow N_2+3H_2$

Now we have to determine the partial pressure of $N_2$ and $H_2$

$p_{N_2}=\frac{n_{N_2}}{n_T}\times p_T$

Given:

$n_{N_2}=1\\\\n_{H_2}=3\\\\n_{T}=4\\\\p_T=866mmHg$

$p_{N_2}=\frac{1}{4}\times (866mmHg)=216.5mmHg$

and,

$p_{H_2}=\frac{n_{H_2}}{n_T}\times p_T$

Given:

$n_{H_2}=1\\\\n_{H_2}=3\\\\n_{T}=4\\\\p_T=866mmHg$

$p_{H_2}=\frac{3}{4}\times (866mmHg)=649.5mmHg$

Thus, the partial pressure of $N_2$ and $H_2$ is, 216.5 mmHg and 649.5 mmHg

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3 years ago

Determine the molality of a solution of methanol dissolved in ethanol for which the mole fraction of methanol is 0.135. Give you

Alja [10]

Answer: The molality of the solution is 0.11 m

Explanation:

We are given:

Mole fraction of methanol = 0.135

This means that 0.135 moles of methanol is present in 1 mole of a solution

Moles of ethanol = 1 - 0.135 = 0.865 moles

To calculate the mass for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of ethanol = 0.865 moles

Molar mass of ethanol = 46 g/mol

$0.865mol=\frac{\text{Mass of ethanol}}{46g/mol}\\\\\text{Mass of ethanol}=(0.865mol\times 46g/mol}=39.79g$

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 (methanol) = 0.135 g

$M_{solute}$ = Molar mass of solute (methanol) = 32 g/mol

$W_{solvent}$ = Mass of solvent (ethanol) = 39.79 g

Putting values in above equation, we get:

$\text{Molality of methanol}=\frac{0.135\times 1000}{32\times 39.79}\\\\\text{Molality of methanol}=0.106m\approx 0.11m$

Hence, the molality of the solution is 0.11 m

6 0

3 years ago

In each of the following blanks, only enter a numerical value.

tatiyna

Answer:

1) 1,... 2

2) 18

3) n= 3 and I=1

Explanation:

1) when l= 0, its an s-sub-level, and only 1 orbital is possible which can carry only 2-electrons

2) the maximum number of electron is given by 2n^2= 2×3^2= 18

3) in 3p, the coefficient of p is the value of n= 3 and l-value of P is 1

5 0

2 years ago