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Dimas [21]
3 years ago
5

A solution is made by dissolving 85.0 mL of sodium fluoride to water until the final volume is 350.0 mL. Identify the solute in

this solution.
A. Sodium flouride
B. Water
C. Sodium fluoride and water
D. Unable to determine
Chemistry
1 answer:
Natasha_Volkova [10]3 years ago
8 0

Answer:

  • <u><em>Sodium fluoride</em></u>

Explanation:

A <em>solution </em>is constituted by two parts: the solute and the solvent (there may be more than 1 solute and more than 1 solvent, but that is not the key of the answer).

<em>Solute</em> is the substace that is dissolved and it is in less amount than the solvent.

<em>Solvent</em> is the substance that can dissolve the solute and it is a greater amount than the solute.

In the given solution:

  • <em>sodium fluoride</em>, <em>85.0 mL</em>, is the solute,
  • <em>water</em>, <em>350.0 mL</em> is the solvent, and
  • the mixture of both substances is the solution.
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noname [10]

Answer:

78.2 g/mol  

Step-by-step explanation:

We can use the <em>Ideal Gas Law</em> to solve this problem:

       pV = nRT

Since n = m/M, the equation becomes

      pV = (m/M)RT     Multiply each side by M

   pVM = mRT               Divide each side by pV

        M = (mRT)/(pV)

Data:

ρ = 2.50 g/L

R = 0.082 16 L·atm·K⁻¹mol⁻¹

T =98 °C

p = 740 mmHg

Calculation:

(a)<em> Convert temperature to kelvins </em>

T = (98 + 273.15) = 371.15 K

(b) <em>Convert pressure to atmospheres </em>

p = 740 × 1/760 =0.9737 atm

(c) <em>Calculate the molar mass </em>

Assume V = 1 L.

   Then m = 2.50 g

            M = (2.50 × 0.082 06 × 371.15)/(0.9737 × 1)

                = 76.14/0.9737

                = 78.2 g/mol

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When 0.5141 g of biphenyl (C12H10) undergoes combustion in a bomb calorimeter, the temperature rises from 25.823 °C to 29.419 °C
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Answer:

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q_{cal}= C \times \Delta T

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q_{rxn}= -q_{cal}= -21.076\,kJ

\Delta_{r}U= -21.076 \times \frac{154}{0.5141}= -6313\, kJ/mol

Therefore, \Delta_{r}U of the reaction is -6313 kJ/mol.

The chemical reaction in bomb calorimeter  is as follows.

C_{12}H_{10}(s)+\frac{27}{2}O_{2}(g)\rightarrow 12CO_{2}(g)+5H_{2}O(g)

Number\,of\,moles\Delta n=(12+5)-\frac{27}{2}=3.5

\Delta_{r}H=\Delta E+ \Delta n. RT

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Therefore, \Delta_{r}H of the reaction is -6312 kJ/mol.

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