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2 years ago

A 4,000 mL solution of AgNO3 contains 17.00 g of solute in water. Calculate the molarity (molar concentration) of the solution.

Chemistry

2 answers:

Ahat [919]2 years ago

8 0

Answer: The molarity of the solution is 0.025 M.

Explanation:

Molarity is the unit of concentration. It is defined as the number of moles of solute that are present in one liter of solution.

Equation used to calculate molarity of the solution is:

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

We are given:

Mass of silver nitrate = 17.00 g

Molar mass of silver nitrate = 170 g/mole

Volume of solution = 4000 mL

Putting values in above equation, we get:

$\text{Molarity of the solution}=\frac{17\times 1000}{170\times 4000}=0.025M$

Hence, the molarity of the solution is 0.025 M

Hitman42 [59]2 years ago

3 0

Moles of AgNO3 = 17/107.86 + 14 + 16x3 = 17/169.86 = 0.1moles. Molar concentration = moles of AgNO3 /volume of solution = 0.1/4 = 0.025 moles/litre.

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its is 1.hydrogen 2. oxygen 3. argon 4. carbon

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2 years ago

When an electron in a 2p orbital of a particular atom makes a transition to the 2s orbital, a photon of approximate wavelength 6

Mariulka [41]

Answer:

The energy difference between these 2p and 2s orbitals is $3.07\times 10^{-19} J$

Explanation:

Wavelength of the photon emitted = $\lambda =646.3 nm =646.3\times 10^{-9} m$

Energy of the photon will corresponds to the energy difference between 2p and 2s orbital = E

Energy of the photon is given by Planck's equation:

$E=\frac{hc}{\lambda }$

h = Planck's constant = $6.626\tiomes 10^{-34} Js$

c = Speed of the light = $3\times 10^8 m/s$

$E=\frac{6.626\tiomes 10^{-34} Js\times 3\times 10^8 m/s}{646.3\times 10^{-9} m}$

$E=3.07\times 10^{-19} J$

The energy difference between these 2p and 2s orbitals is $3.07\times 10^{-19} J$

3 0

2 years ago

The activation energy for the reaction no2(g)+co(g)⟶no(g)+co2(g) is ea = 75 kj/mol and the change in enthalpy for the reaction i

Nonamiya [84]

Answer: 350 kj/mol

Explanation:

As shown below this expression gives the activation energy of the reverse reaction:

EA reverse reaction = EA forward reaction + | enthalpy change |

1) The activation energy, EA is the difference between the potential energies of the reactants and the transition state:

EA = energy of the transition state - energy of the reactants.

2) The activation energy of the forward reaction given is:

EA = energy of the transition state - energy of [ NO2(g) + CO(g) ] = 75 kj/mol

3) The negative enthalpy change - 275 kj / mol for the forward reaction means that the products are below in the potential energy diagram, and that the potential energy of the products, [NO(g) + CO2(g) ] is equal to 75 kj / mol - 275 kj / mol = - 200 kj/mol

4) For the reverse reaction the reactants are [NO(g) + CO2(g)], and the transition state is the same than that for the forward reaction.

5) The difference of energy between the transition state and the potential energy of [NO(g) + CO2(g) ] will be the absolute value of the change of enthalpy plus the activation energy for the forward reaction:

EA reverse reaction = EA forward reaction + | enthalpy change |

EA reverse reaction = 75 kj / mol + |-275 kj/mol | = 75 kj/mol + 275 kj/mol = 350 kj/mol.

And that is the answer, 350 kj/mol

3 0

2 years ago

To determine the concentration of ethanol in cognac a 5.00 mL sample of the cognac is diluted to 0.500 L. Analysis of the dilute

julia-pushkina [17]

Answer : The molar concentration of ethanol in the undiluted cognac is 8.44 M

Explanation :

Using neutralization law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molar concentration of undiluted cognac = ?

$M_2$ = molar concentration of diluted cognac = 0.0844 M

$V_1$ = volume of undiluted cognac = 5.00 mL = 0.005 L

$V_2$ = volume of diluted cognac = 0.500 L

Now put all the given values in the above law, we get molar concentration of ethanol in the undiluted cognac.

$M_1\times 0.005L=0.0844M\times 0.500L$

$M_1=8.44M$

Therefore, the molar concentration of ethanol in the undiluted cognac is 8.44 M

4 0

2 years ago

Phosphorous acid, H3PO3(aq), is a diprotic oxyacid that is an important compound in industry and agriculture. K pKa1 K pKa2 1.30

FrozenT [24]

Answer:

* Before addition of any KOH:

pH = 0,0301

*After addition of 25.0 mL KOH:

pH = 1,30

*After addition of 50.0 mL KOH:

pH = 2,87

*After addition of 75.0 mL KOH:

pH = 6,70

*After addition of 100.0 mL KOH:

pH = 10,7

Explanation:

H₃PO₃ has the following equilibriums:

H₃PO₃ ⇄ H₂PO₃⁻ H⁺

k = [H₂PO₃⁻] [H⁺] / [H₃PO₃] k = 10^-(1,30) (1)

H₂PO₃⁻ ⇄ HPO₃²⁻ + H⁺

k = [HPO₃²⁻] [H⁺] / [H₂PO₃⁻] k = 10^-(6,70) (2)

Moles of H₃PO₃ are:

0,0500L×(1,8mol/L) = 0,09 moles of H₃PO₃

* Before addition of any KOH:

Using (1), moles in equilibrium are:

H₃PO₃: 0,09-x

H₂PO₃⁻: x

H⁺: x

Replacing:

$10^{-1.30} = \frac{x^2}{0.09-x}$

4.51x10⁻³ - 0.050x -x² = 0

The right solution of x is:

x = 0.0466589

As volume is 0,050L

[H⁺] = 0.0466589moles / 0,050L = 0,933M

As pH = -log [H⁺]

pH = 0,0301

*After addition of 25.0 mL KOH:

0,025L×1,8M = 0,045 moles of KOH that reacts with H₃PO₃ thus:

KOH + H₃PO₃ → H₂PO₃⁻ + H₂O

That means moles of KOH will be the same of H₂PO₃⁻ and moles of H₃PO₃ are 0,09moles - 0,045moles = 0,045moles

Henderson-Hasselbalch formula is:

pH = pka + log₁₀ [A⁻] /[HA]

Where A⁻ is H₂PO₃⁻ and HA is H₃PO₃.

Replacing:

pH = 1,30 + log₁₀ [0,045mol] / [0,045mol]

pH = 1,30

*After addition of 50.0 mL KOH:

The addition of 50.0 mL KOH consume all H₃PO₃. Thus, in the solution you will have just H₂PO₃⁻. Thus, moles in solution for the equilibrium will be:

H₂PO₃⁻: 0,09-x

HPO₃²⁻: x

H⁺: x

Replacing:

$10^{-6.70} = \frac{x^2}{0.09-x}$

1.8x10⁻⁸ - 2x10⁻⁷x - x² = 0

The right solution of x is:

x = 0.000134064

As volume is 50,0mL + 50,0mL = 100,0mL

[H⁺] = 0.000134064moles / 0,100L = 1.34x10⁻³M

As pH = -log [H⁺]

pH = 2,87

*After addition of 75.0 mL KOH:

Applying Henderson-Hasselbalch formula you will have 0,045 moles of both H₂PO₃⁻ HPO₃²⁻ and pka: 6,70:

pH = 6,70 + log₁₀ [0,045mol] / [0,045mol]

pH = 6,70

*After addition of 100.0 mL KOH:

You will have just 0,09moles of HPO₃²⁻, the equilibrium will be:

HPO₃²⁻ + H₂O ⇄ H₂PO₃⁻ + OH⁻ with kb = kw/ka = 1x10⁻¹⁴/10^-(6,70) = 5,01x10⁻⁸

kb = [H₂PO₃⁻] [OH⁻] / [HPO₃²⁻]

Moles are:

H₂PO₃⁻: x

OH⁻: x

HPO₃²⁻: 0,09-x

Replacing:

$5.01x10^{-8} = \frac{x^2}{0.09-x}$

4.5x10⁻⁹ - 5.01x10⁻⁸x - x² = 0

The right solution of x is:

x = 0.000067057

As volume is 50,0mL + 100,0mL = 150,0mL

[OH⁻] = 0.000067057moles / 0,150L = 4.47x10⁻⁴M

As pH = 14-pOH; pOH = -log [OH⁻]

pH = 10,7

I hope it helps!

6 0

2 years ago