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tankabanditka [31]

4 years ago

10

What is the [OH-] of a solution prepared by dissolving 0.0912 g of hydrogen chloride in sufficient pure water to prepare 250.0 m

l of solution

1 answer:

velikii [3]4 years ago

8 0

Answer: The $[OH^-]$ of a solution is $10^{-12}$ M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in ml

moles of $HCl$ = $\frac{\text {given mass}}{\text {Molar mass}}=\frac{0.0912g}{36.5g/mol}=0.0025mol$

Now put all the given values in the formula of molality, we get

$Molarity=\frac{0.0025\times 1000}{250}=0.01$

pH or pOH is the measure of acidity or alkalinity of a solution.

$HCl\rightarrow H^++Cl^{-}$

According to stoichiometry,

1 mole of $HCl$ gives 1 mole of $H^+$

Thus $0.01$ moles of $HCl$ gives = $\frac{1}{1}\times 0.01=0.01$ moles of $H^+$

Putting in the values:

$[H^+][OH^-]=10^{-14}$

$[0.01][OH^-]=10^{-14}$

$[OH^-]=10^{-12}$

Thus the $[OH^-]$ of a solution prepared by dissolving 0.0912 g of hydrogen chloride in sufficient pure water to prepare 250.0 ml of solution is $10^{-12}$ M

You might be interested in

If this is a p1000 micropipette, then this is set to dispense [ Select]ul. If this is a p10

mariarad [96]

Answer:

1000 µL; 10 µL

Explanation:

A p1000 micropipet is set to dispense 1000 µL.

A p10 micropipet set to dispense 10 µL.

3 0

4 years ago

A piece of metal with a mass of 114 g was placed into a graduated cylinder that contained 25. 00 ml of water, raising the water

alexgriva [62]

The density of the metal is 6.51 g/cm3 .

<h3>What is density, for instance?</h3>

How much "stuff" is contained in a specific quantity of space is determined by its density. For instance, a block of the harder, lighter material gold (Au) will be denser than a block of the heaviest element lead (Pb) (Au). Styrofoam blocks are less dense than bricks. Mass per unit size serves as its definition.

Briefing :

mass, m = 114 g

initial volume, V1 = 25 mL

final volume, V2 = 42.5 mL

Volume of the metal piece, V = V2 - V1

= 42.5 - 25

= 17.5 mL

1 mL = 1 c.c

So, Volume of metal, V = 17.5 c.c.

Let the density of the metal is d.

density = mass / volume

d = 114 / 17.5

= 6.51 g/c.c

Thus, the density of metal is 6.51 g/c.c.

To know more about density visit ;

brainly.com/question/15164682

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3 0

1 year ago

Calculate the solubilities of the following compounds in a 0.02 M solution of barium nitrate using molar concentrations, first i

Law Incorporation [45]

Answer:

a. 1.7 × 10⁻⁴ mol·L⁻¹; b. 5.5 × 10⁻⁹ mol·L⁻¹

c. 2.3 × 10⁻⁴ mol·L⁻¹; 5.5 × 10⁻⁸ mol·L⁻¹

Explanation:

a. Silver iodate

Let s = the molar solubility.

AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq); Ksp = 3.0 × 10⁻⁸

E/mol·L⁻¹: s s

$K_{sp} =\text{[Ag$^{+}$][IO$_{3}$$^{-}$]} = s\times s = s^{2} = 3.0\times 10^{-8}\\s = \sqrt{3.0\times 10^{-8}} \text{ mol/L} = 1.7 \times 10^{-4} \text{ mol/L}$

b. Barium sulfate

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq); Ksp = 1.1 × 10⁻¹⁰

I/mol·L⁻¹: 0.02 0

C/mol·L⁻¹: +s +s

E/mol·L⁻¹: 0.02 + s s

$K_{sp} =\text{[Ba$^{2+}$][SO$_{4}$$^{2-}$]} = (0.02 + s) \times s \approx 0.02s = 1.1\times 10^{-10}\\s = \dfrac{1.1\times 10^{-10}}{0.02} \text{ mol/L} = 5.5 \times 10^{-9} \text{ mol/L}$

c. Using ionic strength and activities

(i) Calculate the ionic strength of 0.02 mol·L⁻¹ Ba(NO₃)₂

The formula for ionic strength is

$\mu = \dfrac{1}{2} \sum_{i} {c_{i}z_{i}^{2}}\\\\\mu = \dfrac{1}{2} (\text{[Ba$^{2+}$]}\cdot (2+)^{2} + \text{[NO$_{3}$$^{-}$]}\times(-1)^{2}) = \dfrac{1}{2} (\text{0.02}\times 4 + \text{0.04}\times1)= \dfrac{1}{2} (0.08 + 0.04)\\\\= \dfrac{1}{2} \times0.12 = 0.06$

(ii) Silver iodate

a. Calculate the activity coefficients of the ions

$\log \gamma = -0.51z^{2}\sqrt{I} = -0.051(1)^{2}\sqrt{0.06} = -0.51\times 0.24 = -0.12\\\gamma = 10^{-0.12} = 0.75$

b. Calculate the solubility

AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq)

$K_{sp} =\text{[Ag$^{+}$]$\gamma_{Ag^{+}}$[IO$_{3}$$^{-}$]$\gamma_{IO_{3}^{-}}$} = s\times0.75\times s \times 0.75 =0.56s^{2}= 3.0 \times 10^{-8}\\s^{2} = \dfrac{3.0 \times 10^{-8}}{0.56} = 5.3 \times 10^{-8}\\\\s =2.3 \times 10^{-4}\text{ mol/L}$

(iii) Barium sulfate

a. Calculate the activity coefficients of the ions

$\log \gamma = -0.51z^{2}\sqrt{I} = -0.051(2)^{2}\sqrt{0.06} = -0.51\times16\times 0.24 = -0.50\\\gamma = 10^{-0.50} = 0.32$

b. Calculate the solubility

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq

$K_{sp} =\text{[Ba$^{2+}$]$\gamma_{ Ba^{2+}}$[SO$_{4}$$^{2-}$]$\gamma_{ SO_{4}^{2-}}$} = (0.02 + s) \times 0.32\times s\times 0.32 \approx 0.02\times0.10s\\2.0\times 10^{-3}s = 1.1 \times 10^{-10}\\s = \dfrac{1.1\times 10^{-10}}{2.0 \times 10^{-3}} \text{ mol/L} = 5.5 \times 10^{-8} \text{ mol/L}$

7 0

3 years ago

What is the<br> mass of 1.82<br> moles of<br> Lithium<br> carbonate?

frutty [35]

Answer:

244.76

Explanation:

The weight of grams by 1.82 moles of lithium carbonate would be: 134.481438 So we need to use this equation to find mass by grams m × g = ms Where m is for moles, g for grams, and ms for mass. So now we need use this equation: 1.82 × 134.481438 = ? 1.82 × 134.481438 = 244.75621716 244.75621716, rounded-up (to the nearest-tenths place) is 244.76. So now you have it! The mass of 1.82 moles of lithium carbonate is 244.76!

7 0

3 years ago

The masses of carbon and hydrogen in samples of four pure hydrocarbons are given above. The hydrocarbon in which sample has the

enot [183]

Answer:

Sample B

Explanation:

In this case, we need to determine the empirical formula for each sample. The one that match the formula of the propene would be the sample.

Let's do Sample A:

C: 60 g; H: 12 g

1. Calculate moles:

We need the atomic weights of carbon (12 g/mol) and hydrogen (1 g/mol):

C: 60 / 12 = 5

H: 12 / 1 = 12

2. Determine number of atoms in the formula

In this case, we just divide the lowest moles obtained in the previous part, by all the moles:

C: 5 / 5 = 1

H: 12 / 5 = 2.4 or rounded to two

3. Write the empirical formula:

Now, the prior results, represent the number of atoms in the empirical formula for each element, so, we put them with the symbol and the atoms as subscripted:

C₁H₂ = CH₂

Therefore, sample A is not the same as propene.

Sample B:

C: 72 g H: 12 g

Following the same steps, let's determine the empirical formula for this sample

C: 72 / 12 = 6 ---> 6 / 6 = 1

H: 12 / 1 = 12 ----> 12 / 6 = 2

EF: CH₂

Sample C:

C: 84 g H: 10 g

C: 84 / 12 = 7 ----> 7 / 7 = 1

H: 10 / 1 = 10 ----> 10 / 7 = 1.4 or just 1

EF: CH

Sample D

C: 90 g H: 10 g

C: 90 / 12 = 7.5 -----> 7.5 / 7.5 = 1

H: 10 / 1 = 10 -------> 10 / 7.5 = 1.33 or just 1

EF: CH

Neither compound has the same empirical formula as C3H6, but C3H6 is a molecular formula, so, if we just simplify the formula we have:

C3H6 -----> CH₂

Therefore, sample B is the one that match completely. Sample B would be the one.

Hope this helps

8 0

3 years ago