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love history [14]
3 years ago
5

Dimethyl sulfoxide [(ch3)2so], also called dmso, is an important solvent that penetrates the skin, enabling it to be used as a t

opical drug-delivery agent. calculate the number of c, s, h, and o atoms in 7.14 × 103 g of dimethyl sulfoxide.
Chemistry
1 answer:
Bas_tet [7]3 years ago
4 0

The molecular formula of dimethyl sulfoxide is (CH_{3})_{2}SO. Molar mass of dimethyl sulfoxide is 78.13 g/mol. Calculate number of moles as follows:

n=\frac{m}{M}=\frac{7.14\times 10^{3} g}{78.13 g/mol}=91.38 mol

From the molecular formula, 1 mole of dimethyl sulfoxide contains 2 moles of Carbon, 6 moles of Hydrogen, 1 mole of Sulfur and 1 mole of oxygen.

Thus, 91.38 moles of dimethyl sulfoxide will have:

Carbon :

n_{C}=2\times 91.38 moles=182.77 moles

Hydrogen:

n_{H}=6\times 91.38 moles=548.28 moles

Sulfur:

n_{S}=1\times 91.38 moles=91.38 moles

Oxygen:

n_{O}=1\times 91.38 moles=91.38 moles

Since, 1 mole of an element equals to 6.023\times 10^{23} atoms thus, number of atoms can be calculated as:

Carbon:

1 mole\rightarrow 6.023\times 10^{23} atoms

Thus,

182.77 moles\rightarrow 182.77\times 6.023\times 10^{23} atoms=1.10\times 10^{26} atoms

Hydrogen:

1 mole\rightarrow 6.023\times 10^{23} atoms

Thus,

548.28 moles\rightarrow 548.28\times 6.023\times 10^{23} atoms=3.30\times 10^{26} atoms

Sulfur:

1 mole\rightarrow 6.023\times 10^{23} atoms

Thus,

91.38 moles\rightarrow 91.38\times 6.023\times 10^{23} atoms=5.50\times 10^{25} atoms

Oxygen:

1 mole\rightarrow 6.023\times 10^{23} atoms

Thus,

91.38 moles\rightarrow 91.38\times 6.023\times 10^{23} atoms=5.50\times 10^{25} atoms

Therefore, number of C, S, H and O atoms are 1.10\times 10^{26}, 5.50\times 10^{25}, 3.30\times 10^{26} and 5.50\times 10^{25} atoms respectively.

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300 moles of sodium nitrite are needed for a reaction. the solution is 0.450 m. how many ml are needed?
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\boxed{\text{667 000 mL}}

Explanation:

\text{Molar concentration} = \dfrac{\text{moles}}{\text{litres}}\\\\c = \dfrac{n}{v}

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Calculation:

\begin{array}{rcl}0.450 & = & \dfrac{300}{V}\\\\0.450V& = & 300\\\\V & = & \dfrac{300}{0.450}\\\\V & = & \text{667 L} =\textbf{667 000 mL}\end{array}\\\text{The volume of solution needed is }\boxed{\textbf{667 000 mL}}

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