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2 years ago

Calculate the mass percent of carbon (c) in carbon tetrachloride, (ccl4), if the molar mass of carbon tetrachloride is 153.8g.

Chemistry

2 answers:

baherus [9]2 years ago

3 0

mass percent of carbon (c) in CCl₄ = mass of C/mass of CCl₄
= 12 x 100/153.8 = 7.8%

NeTakaya2 years ago

3 0

Answer: The mass percent of carbon in carbon tetrachloride is 7.8 %.

Explanation:

In $CCl_4$ , there are 1 carbon atoms and 4 chlorine atoms.

To calculate the mass percent of element in a given compound, we use the formula:

$\text{Mass percent of carbon}=\frac{\text{Mass of carbon}}{\text{Molar mass of carbon tetrachloride}}\times 100$

Mass of carbon = 12 g

Mass of carbon tetrachloride = 153.8 g

Putting values in above equation, we get:

$\text{Mass percent of carbon}=\frac{12}{153.8}\times 100=7.80\%$

Hence, the mass percent of carbon in carbon tetrachloride is 7.8 %.

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Which of the following statements is true about the following reaction?

MAVERICK [17]

The correct reaction equation is:

$3NaHCO_{3} (aq) + C_{6}H_{8}O_{7} (aq) \rightarrow 3CO_{2} (g) + 3H_{2}O (l) + Na_{3}C_{6}H_{5}O_{7} (aq)$

Answer:

b) 1 mole of water is produced for every mole of carbon dioxide produced.

Explanation: CONVERT EVERYTHING TO MOLES OR VOLUME, THEN COMPARE IT WITH THE COMPOUND'S STOICHIOMETRY IN CHEMICAL EQUATION.

a) 22.4 L of $CO_{2}$ gas is produced only when $\frac{22.4}{3}$ L of $C_{6}H_{8}O_{7}$ is reacted with 22.4 L of $NaHCO_{3}$ . So it is wrong.

b) Since in the chemical equation the stoichiometric coefficient of $CO_{2}$ and $H_{2}O$ are same so the number of moles or volume of each of them will be same whatever the amount of reactants taken. Therefore it is correct option.

c) $6.02\times 10^{23}$ molecules is equal 1 mole of $Na_{3}C_{6}H_{5}O_{7}$ if produced then 3 moles of $NaHCO_{3}$ is required, which is not given in the option. So it is wrong.

d) 54 g of water or 3 moles of $H_{2}O$ (Molecular Weight of water is 18 g) is produced when 3 moles of $NaHCO_{3}$ is used but in this option only one mole of $NaHCO_{3}$ is given. So it is wrong.

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2 years ago

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Plzzz help...

Anna [14]

Answer:

just answer this and you will have yours

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1 year ago

If you were to assume a 95% yield for the formation of the phosphonium ion, how many milligrams of benzyl chloride and triphenyl

luda_lava [24]

I am attempting the problem for phosphonium Ion rather than its chloride salt. The chemical equation is shown below along with molar masses in mg.

First of all we will calculate the amounts of reactants required for the synthesis of 220 mg of phophonium ion. Calculations for both reactants is as follow,

For Benzyl chloride,
$\frac{353420 mg of Phosphonium is formed by reacting}{220 mg of Phosphonium will be formed by}$ = $\frac{126580 g of benzyl chloride}{X}$

Solving for X,
X = $\frac{220 mg . 126580 mg}{353420 mg}$
X = 78.79 mg

For PPh₃:
$\frac{353420 mg of Phosphonium is formed by reacting}{220 mg of Phosphonium will be formed by}$ = $\frac{262290 g of PPh3}{X}$

Solving for X,
X = $\frac{220 mg . 262290 mg}{353420 mg}$
X = 163.27 mg

Now, Assuming these values as for 95 % conversion, we can calculate 100 % yield as follow,

when $\frac{95 percent}{100 percent}$ = $\frac{220 g}{X}$
Solving for X,

X = $\frac{220 . 100}{95}$ = 231.57 mg

Now, calculate reactants mass with respect to 231.57 mg
when $\frac{220 mg phosphonium required}{231.57 mg require}$ = $\frac{78.79 g of benzyl chloride}{X}$
Solving for ,

X = $\frac{231.57 . 78.79}{220}$ = 82.93 mg of Benzyl chloride

when $\frac{220 mg phosphonium required}{231.57 mg require}$ = $\frac{163.27 g of PPh3}{X}$
Solving for ,

X = $\frac{231.57 . 163.27}{220}$ = 171.85 mg of PPh3

So, reaction was started with reacting 82.93 mg of Benzyl Chloride and 171.85 mg of Triphenyl Phosphine.

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2 years ago

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bonufazy [111]

Answer:remove 4 electrons

Explanation:

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2 years ago

If a chemist analyzes a 3.84g sample containing sand and table sugar, and recovers 1.43g of sand, what percent by mass of

disa [49]

3.84 - 1.43 = 2.41
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2 years ago

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