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3 years ago

Solution X has a pH of 9.4, and solution Y has a pH of 7.4.

Chemistry

2 answers:

vodomira [7]3 years ago

8 0

Answer:

a. [H30+] = 4 * 10^-10 M

b. [H3O+]= 4 * 10^-8 M

Explanation:

Step 1: Data given

pH of solution X = 9.4

pH of solution Y = 7.4

Step 2: Calculate [H3O+] of solution X

pH = -log [H+]

[H+] = [H30+]

pH = -log[H3O+]

9.4 = -log[H3O+]

10^-9.4 = 3.98*10*-10

[H30+] = 4 * 10^-10 M

To control we can calculate pH

-log[4 * 10^-10] = 9.4

Step 3: Calculate [H3O+] of solution Y

pH = -log [H+]

[H+] = [H30+]

pH = -log[H3O+]

7.4 = -log[H3O+]

10^-7.4 = 4*10*-8

[H30+] = 4 * 10^-8 M

To control we can calculate pH

-log[4 * 10^-8] = 74

a. [H30+] = 4 * 10^-10 M

b. [H3O+]= 4 * 10^-8 M

Whitepunk [10]3 years ago

4 0

Answer:

see below

Explanation:

[H₃O⁺] = 10^-pH

a) Soln X with pH = 4.9 => [H₃O⁺] = 10⁻⁴°⁹ = 2.51 x 10⁻⁵M

b) Soln Y with pH = 7.4 => [H₃O⁺] = 10⁻⁷°⁴ = 3.98 x 10⁻⁸M

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Complete end balance the reaction :copper plus hydrobromic acid

Reika [66]

Answer:

The correct answer is - 2Cu + 4HBr → 2H[CuBr2] + H2

Explanation:

Copper (Cu) and hydrobromic acid (HBr), which is a dissolved form of hydrogen bromide in the water, react and produce hydrogen dibromidocuprate (I) and hydrogen molecules.

The complete and balanced equation is -

2Cu + 4HBr → 2H[CuBr2] + H2

This reaction takes place in the medium of ester. Copper is in the suspension form in this reaction.

Thus, the correct answer is - 2Cu + 4HBr → 2H[CuBr2] + H2

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3 years ago

What is the empirical formula for a compound that is 83.7% carbon and 16.3% hydrogen?

zubka84 [21]

Hello!

We use the amount in grams (mass ratio) based on the composition of the elements, see: (in 100 g solution)

C: 83.7% = 83,7 g
H: 16.3% = 16.3 g

Let us use the above mentioned data (in g) and values will be converted to amount of substance (number of moles) by dividing by molecular mass (g / mol) each of the values, lets see:

$C: \dfrac{83.7\:\diagup\!\!\!\!\!g}{12\:\diagup\!\!\!\!\!g/mol} \approx 6.975\:mol$

$H: \dfrac{16.3\:\diagup\!\!\!\!\!g}{1\:\diagup\!\!\!\!\!g/mol} = 16.3\:mol$

We note that the values found above are not integers, so let's divide these values by the smallest of them, so that the proportion is not changed, let's see:

$C: \dfrac{6.975}{6.975} = 1$

$H: \dfrac{16.3}{6.975} \approx 2.3$

Note: So the ratio in the smallest whole numbers of carbon to hydrogen is 3:7, t<span>hus, the minimum or empirical formula found for the compound will be:
</span>
$\boxed{\boxed{C_3H_7}}\end{array}}\qquad\checkmark$

I hope this helps. =)

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3 years ago

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Vilka [71]

The period is the end of the sentence!!!

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3 years ago

What happened to the molecule when the electron was promoted to the antibonding orbital?.

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the bond will break

The bond will dissolve (break) if the electron absorbs a photon and is moved from a bonding molecular orbital to an antibonding orbital since there is no longer an overall stabilizing interaction.

<h3>What is an antibonding orbital?</h3>

An antibonding molecular orbital is the molecular orbital created by the destructive overlapping of atomic orbitals.

<h3>Why is it called antibonding orbital?</h3>

Every atom will add one electron to the bond that makes up the lower energy bond.
To prevent interacting with the other two electrons, the additional electron will occupy a higher energy state.
The antibonding orbital is the name of this higher energy orbital.

<h3>What orbitals form an antibond?</h3>

The bonding orbitals are home to electrons that spend the majority of their time between the nuclei of two atoms, whereas the antibonding orbitals are home to electrons that spend the majority of their time outside the nuclei of two atoms.

<h3>When an electron was elevated to the antibonding orbital, what happened?</h3>

In contrast, putting electrons in antibonding orbitals will make the molecule less stable.
The energy levels of the orbitals will determine how many electrons are filled.
The lower energy orbitals will be filled first, and then the higher energy orbitals.

<h3 />

To learn more about antibonding orbitals visit:

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4 0

2 years ago

1.00 M CaCl2 Density = 1.07 g/mL

Lesechka [4]

Explanation:

Molarity of solution = 1.00 M = 1.00 mol/L

In 1 L of solution 1.00 moles of calcium chloride is present.

Mass of solute or calcium chloride = m

$m = 1 mol\times 111 g/mol = 111 g$

Mass of solution = M

Volume of solution = V = 1L = 1000 mL

Density of solution , d= 1.07 g/mL

$M=d\times V=1.07 g/mL\times 1000 mL=1,070 g$

1) The value of %(m/M):

$\frac{m}{M}\times 100=\frac{111 g}{1,070 g}\times 100=10.37\%$

2) The value of %(m/V):

$\frac{m}{V}\times 100=\frac{111 g}{1000 L}\times 100=11.1\%$

$Molality = \frac{\text{Moles of compound }}{\text{mass of solvent in kg}}$

$Normality=\frac{\text{Moles of compound }}{n\times \text{volume of solution in L}}$

n = Equivalent mass

n = $\frac{\text{molar mass of ion}}{\text{charge on an ion}}$

3) Normality of calcium ions:

Moles of calcium ion = 1 mol (1 $CaCl_2$ mole has 1 mole of calcium ion)

$n=\frac{40 g/mol}{2}=20$

$=\frac{1 mol}{20 g/mol\times 1L}=0.050 N$

4) Normality of chlorine ions:

Moles of chlorine ion = 2 mol (1 $CaCl_2$ mole has 2 mole of chlorine ion)

$n=\frac{35.5 g/mol}{1}=35.5$

$=\frac{2 mol}{35.5 g/mol\times 1L}=0.056 N$

Moles of calcium chloride = 1.00 mol

Mass of solvent = Mass of solution - mass of solute

= 1,070 g - 111 g = 959 g = 0.959 kg ( 1 g =0.001 kg)

5) Molality of the solution :

$\frac{1 mol}{0.959 kg}=1.043 mol/kg$

Moles of calcium chloride = $n_1=1mol$

Mass of solvent = 959 g

Moles of water = $n_2=\frac{959 g}{18 g/mol}=53.28 mol$

Mass of solvent = 959 g

6) Mole fraction of calcium chloride =

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{1mol}{1 mol+53.28 mol}=0.01842$

7) Mole fraction of water =

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{53.28 mol}{1mol+53.28 mol}=0.9816$

8) Mass of solution = m'

Volume of the solution= v = 100 mL

Density of solution = d = 1.07 g/mL

$m'=d\times v=1.07 g/ml\times 100 g= 107 g$

Mass of 100 mL of this solution 107 grams of solution.

9) Volume of solution = V = 100 mL

Mass of solution = M'' = 107 g

Mass of solute = m

The value of %(m/V) of solution = 11.1%

$11.1\%=\frac{m}{100 mL}\times 100$

m = 11.1 g

Mass of solvent = M''- m = 107 g -11.1 g = 95.9 g

95.9 grams of water was present in 100 mL of given solution.

3 0

3 years ago