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3 years ago

Find the percentage composition of a compound that contains 1.94 g of carbon, 0.48 g of hydrogen, and 2.58 g of sulfur.

1 answer:

3 0

Answer : The percentage composition of carbon, hydrogen and sulfur in a compound is, 38.8 %, 9.6 % and 51.6 % respectively.

Explanation :

To calculate the percentage composition of element in sample, we use the equation:

$\%\text{ composition of element}=\frac{\text{Mass of element}}{\text{Mass of sample}}\times 100$

Given:

Mass of carbon = 1.94 g

Mass of hydrogen = 0.48 g

Mass of sulfur = 2.58 g

First we have to calculate the mass of sample.

Mass of sample = Mass of carbon + Mass of hydrogen + Mass of sulfur

Mass of sample = 1.94 + 0.48 + 2.58 = 5.0 g

Now we have to calculate the percentage composition of a compound.

$\%\text{ composition of carbon}=\frac{1.94g}{5.0g}\times 100=38.8\%$

$\%\text{ composition of hydrogen}=\frac{0.48g}{5.0g}\times 100=9.6\%$

$\%\text{ composition of sulfur}=\frac{2.58g}{5.0g}\times 100=51.6\%$

Hence, the percentage composition of carbon, hydrogen and sulfur in a compound is, 38.8 %, 9.6 % and 51.6 % respectively.

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anyanavicka [17]

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you re write the formula and solve for volume

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then you input the values

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n = moles = 0.2540

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T = C degrees + 273.15 = K

solve for voume

make sure all units match

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At top, Image 1 shows 2 C's connected by 2 lines. Each C is connected by single lines to 2 H's. At bottom, two dots and 12 C's i

Nana76 [90]

Answer: c

Explanation: Trust me

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2. Nitric oxide contains 46.66% nitrogen and 53.34% oxygen. Water contains 11.21% hydrogen and 88.79% oxygen. Ammonia contains 1

Mrrafil [7]

Answer:

The law of reciprocal proportions states that if two elements react individually with a given weight of a third element, the ratio of the masses with which they combine with the third element are either the same or a simple multiple of the ratio of the masses with which they combine with each other

The compounds formed includes;

1) Nitric oxide, NO

Nitrogen = 46.66% × 30.01 = 14

Oxygen = 53.34% × 30.01 = 16

2) Water, H₂O

Hydrogen = 11.21% × 18.01528 = 2

Oxygen = 88.79% × 18.01528 ≈ 16

3) Ammonia, NH₃

Hydrogen = 17.78% × 17.031 ≈ 3

Nitrogen = 82.22% × 17.031 ≈ 14

The ratio of nitrogen to oxygen in nitric oxide = 14:16 = 7:8

The ratio of nitrogen to hydrogen in ammonia = 14:3

The ratio in which hydrogen and oxygen combine with nitrogen = 3/16

The ratio of hydrogen and oxygen combine with each other in water = 2/16

Therefore, the ratio with which hydrogen and oxygen combine with nitrogen, is (2/3) times the ratio with which they combine with each other, which verifies the law of reciprocal proportions

Explanation:

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Sliva [168]

Answer : The time passed in years is $2.83\times 10^3\text{ years}$

Explanation :

Half-life = 5730 years

First we have to calculate the rate constant, we use the formula :

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{5730\text{ years}}$

$k=1.21\times 10^{-4}\text{ years}^{-1}$

Now we have to calculate the time passed.

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $1.21\times 10^{-4}\text{ years}^{-1}$

t = time passed by the sample = ?

a = let initial amount of the reactant = X g

a - x = amount left after decay process = $71\% \times (x)=\frac{71}{100}\times (X)=0.71Xg$

Now put all the given values in above equation, we get

$t=\frac{2.303}{1.21\times 10^{-4}}\log\frac{X}{0.71X}$

$t=2831.00\text{ years}=2.83\times 10^3\text{ years}$

Therefore, the time passed in years is $2.83\times 10^3\text{ years}$

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