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2 years ago

A 2400.-gram sample of an aqueous solution contains 0.012 gram of nh3. What is the concentration of nh3 in the solution

Chemistry

1 answer:

Kryger [21]2 years ago

7 0

Answer : The concentration of $NH_3$ in the solution is, $2.94\times 10^{-4}M$

Explanation :

First we have to calculate the volume of aqueous solution that is water.

Density of water = 1.00 g/mL

Mass of water = 2400 g

$\text{Volume of water}=\frac{Mass}{Density}=\frac{2400g}{1.00g/mL}=2400mL$

Now we have to calculate the concentration of ammonia solution.

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

$\text{Molarity}=\frac{\text{Mass of }NH_3\times 1000}{\text{Molar mass of }NH_3\times \text{Volume of solution (in mL)}}$

Molar mass of $NH_3$ = 17 g/mole

Now put all the given values in this formula, we get:

$\text{Molarity}=\frac{0.012g\times 1000}{17g/mole\times 2400mL}=3.12mole/L=2.94\times 10^{-4}M$

Therefore, the concentration of $NH_3$ in the solution is, $2.94\times 10^{-4}M$

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Use average bond energies to calculate ΔHrxn for the following hydrogenation reaction: H2C=CH2(g)+H2(g)→H3C−CH3(g)

marissa [1.9K]

Answer:

The $\Delta H_{rxn}$ of the given reaction is -129.6 kJ

Explanation:

The given chemical reaction is as follows.

$H_{2}C=CH_{2}(g)+H_{2}(g)\rightarrow H_{3}C-CH_{3}(g)$

Enthalpy of each reactant and products are as follows.

$\Delta H_{C=C}\,=615.0\,kJ\,mol^{-1}$

$\Delta H _{H-H}\,=435.1\,kJ\,mol^{-1}$

$\Delta H _{C-C}\,=347.3\,kJ\,mol^{-1}$

$\Delta H _{C-H}\,=416.2\,kJ\,mol^{-1}$

In the given chemical reaction involved two C-H bonds in the reactant side and one C-C bond in the product side therefore, the enthalpy of formation will be the negative.

$\Delta H_{rxn}=-\Delta H_{C-C}-2\Delta H_{C-H}+\Delta H_{C=C}+\Delta H_{H-H}$

$=-347.4-2\times416.2+615.0+435.1$

$=-129.6 \,kJ$

Therefore, The $\Delta H_{rxn}$ of the given reaction is -129.6 kJ

4 0

2 years ago

When 1.04g of cyclopropane was burnt in excess oxygen in a bomb calorimeter, the temperature rose by 3.69K. The total heat capac

STatiana [176]

Answer:

$\Delta _{comb}H=-2,093\frac{kJ}{mol}$

Explanation:

Hello!

In this case, since these calorimetry problems are characterized by the fact that the calorimeter absorbs the heat released by the combustion of the substance, we can write:

$Q_{rxn}+Q_{cal}=0$

Thus, given the temperature change and the total heat capacity, we obtain the following total heat of reaction:

$Q_{rxn}=-14.01kJ/K*3.69K\\\\Q_{rxn}=-51.70kJ$

Now, by dividing by the moles in 1.04 g of cyclopropane (42.09 g/mol) we obtain the enthalpy of combustion of this fuel:

$n=\frac{1.04g}{42.09g/mol}=0.0247mol\\\\\Delta _{comb}H=\frac{Q_{rxn}}{n}\\\\ \Delta _{comb}H=-2,093\frac{kJ}{mol}$

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4 0

1 year ago

What is the mass of a sample of NH3 containing 7.20 × 1024 molecules of NH3? 161 grams 187 grams 203 grams 214 grams

Shkiper50 [21]

Answer:

203 grams

Explanation:

It is known that 1.0 mole of a compound contains Avogadro's number of molecules (6.022 x 10²³).

Using cross multiplication:

1.0 mol contains → 6.022 x 10²³ molecules.

??? mol contains → 7.2 x 10²⁴ molecules.

∴ The no. of moles of (6.3 x 10²⁴ molecules) of NH₃ = (1.0 mol)(7.2 x 10²⁴ molecules)/(6.022 x 10²³ molecules) = 11.96 mol.

∴ The no. of grams of NH₃ present = no. of moles x molar mass = (11.96 mol)(17.0 g/mol) = 203.3 g ≅ 203.0 g.

7 0

2 years ago

Read 2 more answers

What happens to energy when Sally kicks a soccer ball?

Vinvika [58]

Answer:

Kinetic energy is transferred from the leg to the soccer ball.

Explanation:

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2 years ago

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Express each aqueous concentration in the unit indicated.

MAXImum [283]

Answer:

a. ppb of trichloroethylene = 3 × 10⁶ ppb

b. ppm of Cl₂ = 3.8 ppm

c. Molarity = 0.0002 mol / L

d. Molarity = 0.0007 mol / L

e. For trace amount of concentrations

Explanation:

a. Given data

mass of trichloroethylene = 25 mg

Volume of water = 9.5 L

ppb of trichloroethylene = ?

Solution

As we know that

1 L = 1000 milliliters

9.5 L = 9.5 × 1000

9.5 L = 9500 millileters (ml)

we consider 25 mg = 25 millileters

ppb = (mass of solute / mass of solvent) × 1000,000,000 (1 billion)

ppb of trichloroethylene = (25 ÷ 9500) × 1000,000,000

ppb of trichloroethylene = 0.003 × 1000,000,000

ppb of trichloroethylene = 3 × 10⁶ ppb

B. Given data

Mass of Cl₂ = 38 g

volume of water = 1.00 × 10⁴ L ( 10000 L)

ppm of Cl₂ = ?

Solution

Volume of water in ml = 1 L = 1000 ml

Volume of water in ml = 10000 × 1000

Volume of water in ml = 10000000 ml

we take 38 g = 38 ml

Now we convert it to ppm

ppm = (mass of solute / mass of solvent) × 1000000 (1 million)

ppm of Cl₂ = ( 38 ÷ 10000000 ) × 1000000

ppm of Cl₂ = 0.0000038 × 1000000

ppm of Cl₂ = 3.8 ppm

C. Given data

Concentration of F⁻ ( Fluoride ion) = 2.4 ppm

Molarity = ?

Solution

As we know that 1 ppm = 0.001 g / L

2.4 ppm = 2.4 × 0.001 g/L

2.4 ppm = 0.0024 g/L

Mass of flouride ions = 0.0024 g

Now we find number of moles

moles = mass / molar mass

molar mass of F⁻ = 19 g/mol

moles of F⁻ = 0.0024 g / 19 g/mol

moles of F⁻ = 0.0002 mol

Molarity = mol of solute / liter of solution

Molarity = 0.0002 mol / 1 L

Molarity = 0.0002 mol / L

D. Given data

Concentration of NO₃⁻ ( nitrate ion) = 45 ppm

Molarity = ?

Solution

As we know that 1 ppm = 0.001 g / L

45 ppm = 45 × 0.001 g/L

45 ppm = 0.045 g/L

Mass of nitrate ions = 0.045 g

Now we find number of moles

moles = mass / molar mass

molar mass of NO₃⁻ = 62 g/mol

moles of NO₃⁻ = 0.045 g / 62 g/mol

moles of F⁻ = 0.0007 mol

Molarity = mol of solute / liter of solution

Molarity = 0.0007 mol / 1 L

Molarity = 0.0007 mol / L

E. Reason of expressing concentration in ppm and ppb

Scientist prefer ppm and ppb notations when the concentration difference of solute and solvent are very high.

As water contains contaminants is a very low amount we can say in trace amounts so scientist prefer ppm and ppb rather than molarity.

Example

Arcenic is an under ground water contaminant and its concentration of 10 μg/L is dangerous for health.

Lets change this in to molarity

mass = 10 μg

10 μg = 10 / 1000000

10 μg = 0.00001 g

now find out moles of Arcenic

moles = mass / molar mass

molar mass of arcenic = 75 g/mol

moles = mass / molar mass

moles of arcenic = 0.00001 g / 75 g/mol

moles of arcenic = 0.00000012 mol

Molarity = moles of solute / litres of solution

Molarity = 0.00000012 mol / 1 L

Molarity = 0.00000012 mol/ L

As we can see that in molarity it is a negligible amount so scientists express it in ppm and ppb

7 0

2 years ago