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3 years ago

A flask contains 0.450 mol of liquid bromine, br2. determine the number of bromine molecules present in the flask.

Chemistry

1 answer:

Pachacha [2.7K]3 years ago

5 0

So number molecules in the sample = moles x 6.022x10^23 molecules/mole
= 0.450 mol x 6.022x10^23
= 2.71x10^23 molecules

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The vapor pressure of benzene at 298 K is 94.4 mm of Hg. The standard molar Gibbs free energy of formation of liquid benzene at

horsena [70]

Answer:

ΔfG°(C₆H₆(g)) = 129.7kJ/mol

Explanation:

Bringing out the parameters mentioned in the question;

Vapor pressure = 94.4 mm of Hg

The vaporization reaction is given as;

C₆H₆(l) ⇄ C₆H₆(g)

Equilibrium in terms of activities is given by:

K = a(C₆H₆(g)) / a(C₆H₆(l))

Activity of pure substances is one:

a(C₆H₆(l)) = 1

Assuming ideal gas phase activity equals partial pressure divided by total pressure. At standard conditions

K = p(C₆H₆(g)) / p°

Where p° = 1atm = 760mmHg standard pressure

We now have;

K = 94mmHg / 760mmHg = 0.12421

Gibbs free energy is given as;

ΔG = - R·T·ln(K)

where R = gas constant = 8.314472J/molK

So ΔG° of vaporization of benzene is:

ΔvG° = - 8.314472 · 298.15 · ln(0.12421)

ΔvG° = 5171J/mol = 5.2kJ/mol

Gibbs free energy change of reaction = Gibbs free energy of formation of products - Gibbs free energy of formation of reactants:

ΔvG° = ΔfG°(C₆H₆(g)) - ΔfG°(C₆H₆(l))

Hence:

ΔfG°(C₆H₆(g)) = ΔvG°+ ΔfG°(C₆H₆(l))

ΔfG°(C₆H₆(g)) = 5.2kJ/mol + 124.5kJ/mol

ΔfG°(C₆H₆(g)) = 129.7kJ/mol

6 0

3 years ago

What was the eutectic temperature (temperature from the two lines of best fit cross) for the mixture

Ymorist [56]

Answer:

hello your question is incomplete below is the missing part of the question

answer : 104°c

Explanation:

The Eutectic temperature for the mixture is 104°c

From the chart attached below it can be seen that the temperature from the two lines of best fit cross is 104°c

7 0

3 years ago

Identify the right

Basile [38]

.answer:

so the first one is analogous

the second is homologous

and the third is analogous

explanation:

analogous. this means they share a similar function (flight) but do not have the same embryonic origin.

6 0

3 years ago

Calculate the Ka for the following acid. Determine if it is a strong or weak acid.

vladimir1956 [14]

Answer:

a) Ka= 7.1 × 10⁻⁴; This is a weak acid because the acid is not completely dissociated in solution.

Explanation:

Step 1: Write the dissociation reaction for nitrous acid

HNO₂(aq) ⇄ H⁺(aq) and NO₂⁻(aq)

Step 2: Calculate the acid dissociation constant

Ka = [H⁺] × [NO₂⁻] / [HNO₂]

Ka = 0.022 × 0.022 / 0.68

Ka = 7.1 × 10⁻⁴

Step 3: Determine the strength of the acid

Since Ka is very small, nitrous acid is a weak acid, not completely dissociated in solution.

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3 years ago

Nitrogen dioxide is a red-brown gas responsible for the brown color of smog. A container of nitrogen dioxide that is at low pres

____ [38]

Answer:

Initially $1.51\times 10^{-2} moles$ of nitrogen dioxide were in the container .

Explanation:

Volume of the container at low pressure and at room temperature = $V_1=3.4 L$

Number of moles in the container = $n_1$

After more addition of nitrogen gas at the same pressure and temperature.

Volume of the container after addition = $V_2=5.11 L$

Number of moles in the container after addition= $n_2=2.28\times 10^{-2} mol$

Applying Avogadro's law:

$\frac{Volume}{Moles}=constant$ (at constant pressure and temperature)

$\frac{V_1}{n_1}=\frac{V_2}{n_2}$

$n_1=\frac{V_1\times n_2}{V_2}=\frac{3.4 L\times 2.28\times 10^{-2} mol}{5.11 L}$

$n_1=1.51\times 10^{-2} mol$

Initially $1.51\times 10^{-2} moles$ of nitrogen dioxide were in the container .

8 0

3 years ago