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Vitek1552 [10]
3 years ago
6

Crude oil pumped out of the ground may be accompanied by formation water, a solution that contains high concentrations of nacl a

nd other salts. If the boiling point of a sample of formation water is 2.30°c above the boiling point of pure water, what is the molality of particles in the sample?
Chemistry
1 answer:
djverab [1.8K]3 years ago
8 0

solution:

the change in the boiling point is given as,

dTbp =2.30°c

elevation constant for the solvent is given by,

kb=0.512°c/m

molality=\frac{dTBP}{KB\times m}\\=\frac{2.30}{0.512c/m}

= 4.49m

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An example of a general formula of an acid is
<span>HCl

This is called as hydrochloric acid or hydrogen chloride.

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3 years ago
The decomposition of nitryl chloride is described by the following chemical equation: 2NO2C1(g) → 2NO2 (g)-C12 (g) Suppose a two
Gnoma [55]

Answer:

Second reaction

NO2 + F -------> NO2F

Rate of reaction:

k1 [NO2] [F2]

Explanation:

NO2 + F2 -----> NO2F + F slow step1

NO2 + F -------> NO2F fast. Step 2

Since the first step is the slowest step, it is the rate determining step of the reaction

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rate = k1 [NO2] [F2]

6 0
3 years ago
How many are molecules ( or formula) in each sample?
andre [41]

Answer:

  • 4.010 \times 10^{25} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}
  • 16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}

<u>Explanation</u>:

<u>Number of molecules for 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}</u>

\text { Firstly molar mass is calculated of } \mathrm{NaHCO}_{3}:

Atomic mass of Na + H + C + 3(O)  = 22.99 + 1.008 + 12.01 + 3 × 16.00 = 84.00 g/mol

\text { Number of molecules of } \mathrm{NaHCO}_{3} \text { in } 55.93 \text { kg are as follows: }

55.93 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} \mathrm{NaHCO}_{3}}{84.00 \mathrm{gm} \mathrm{NaHCO}_{3}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number }\right)

=4.010 \times 10^{26} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}

<u>Number of molecules for for \left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}</u>

\text { Firstly molar mass is calculated of } \mathrm{Na}_{3} \mathrm{PO}_{4}

= Atomic mass of 3(Na) + P + 4(O)

= 3(22.99) + 30.97 + 4(16.00) = 163.94 g/mol

459 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} N a_{3} P O_{4}}{163.94 \mathrm{gm} N a_{3} P O_{4}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number) } / 1 \mathrm{mol}\right.

=16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}

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