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2 years ago

A solution has a pH of 2.4. What happens to the solution if the hydronium ion concentration is increased?

2 answers:

8 0

If the hydronium ion concentration is increased, the pH of the solution decreases.

The concentration of hydrogen ions in a solution expressed usually in moles per liter or in pH units and used as a measure of the acidity of the solution indicator dyes for narrow ranges of hydrogen-ion concentration

Lelu [443]2 years ago

8 0

Answer:

The pH would decrease and the solution would become more acidic

Explanation:

pH of a solution is a measure of its hydronium ion (H3O+) concentration. Mathematically it is given as:

$pH = -log[H3O+] -------(1)\\\\i.e. [H3O+] = 10^{-pH} ------(2)\\$

It is given that pH of the solution = 2.4

Therefore based on equation (2), the [H3O+] would be:

$[H3O+] = 10^{-2.4} = 3.98*10^{-3} M$

If the [H3O+] concentration were increased then based on equation (1) the pH of the solution would decrease i.e. it would become more acidic.

For example, lets say that the new [H3O+] = 5.0 * 10⁻³ M, then

$pH = -log[5*10^{-3} ] = 2.3$

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Unlike an element, a(n) ________ may be separated through a physical process like filtration.

Amiraneli [1.4K]

The answer is mixture, hope this helps!

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2 years ago

A teacher gives students four liquids commonly found in the kitchen - vinegar, apple juice, dish detergent, and milk - and pH in

VMariaS [17]

Answer:

1. Comparative

2. Independent variable

3. The pH is the Dependent variable

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3 years ago

Read 2 more answers

3.5g of a Certain compound X, known to be made of carbon, hydrogen, and perhaps oxygen, and to have a molecular molar mass of 15

shutvik [7]

Answer:

C₅H₁₀O₅

Explanation:

1. Calculate the mass of each element in 2.78 mg of X.

(a) Mass of C

$\text{Mass of C} = \text{5.13 g CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{1.400 g C}$

(b) Mass of H

$\text{Mass of H} = \text{2.10 g H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g H$_{2}$O}} = \text{0.2349 g H}$

(c) Mass of O

Mass of O = 3.5 - 1.400 - 0.2349 = 1.87 g

2. Calculate the moles of each element

$\text{Moles of C = 1400 mg C}\times\dfrac{\text{1 mmol C}}{\text{12.01 mg C }} = \text{116.6 mmol C}\\\\\text{Moles of H = 234.9 mg H} \times \dfrac{\text{1 mmol H}}{\text{1.008 mg H}} = \text{233.1 mmol H}\\\\\text{Moles of O = 1870 mg O} \times \dfrac{\text{1 mmol O}}{\text{16.00 mg O}} = \text{116 mmol O}$

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{116.6}{116.6}= 1\\\\\text{H: } \dfrac{233.1}{116.6} = 1.999\\\\\text{O: } \dfrac{116}{116.6} = 1.00$

4. Round the ratios to the nearest integer

C:H:O = 1:2:1

5. Write the empirical formula

The empirical formula is CH₂O.

6. Calculate the molecular formula.

EF Mass = (12.01 + 2.016 + 16.00) u = 30.03 u

The molecular formula is an integral multiple of the empirical formula.

MF = (EF)ₙ

$n = \dfrac{\text{MF Mass}}{\text{EF Mass }} = \dfrac{\text{150 u}}{\text{30.03 u}} = 5.00 \approx 5$

MF = (CH₂O)₅ = C₅H₁₀O₅

The molecular formula of X is C₅H₁₀O₅.

8 0

3 years ago

What is the energy of a photon of red light that has a frequency of 4.48 X 10^14Hz?

lana [24]

Answer:

E = 29.7× 10⁻²⁰ j

Explanation:

Given data;

Frequency of light = 4.48 × 10¹⁴ Hz

Energy of photon = ?

Solution:

Formula:

E = h.f

E = energy of photon

h = planck's constant

f = frequency

E = h.f

E = 6.63 × 10⁻³⁴ Kg.m² /s × 4.48 × 10¹⁴ s⁻¹

E = 29.7× 10⁻²⁰ Kg.m²/s²

Kg.m²/s² = j

E = 29.7× 10⁻²⁰ j

3 0

2 years ago

The first-order rate constant for the reaction of methyl chloride (CH3Cl) with water to produce methanol (CH3OH) and hydrochlori

Dvinal [7]

Answer:

K(48.5°C) = 1.017 E-8 s-1

Explanation:

CH3Cl + H2O → CH3OH + HCl

at T1 = 25°C (298 K) ⇒ K1 = 3.32 E-10 s-1

at T2 = 48.5°C (321.5 K) ⇒ K2 = ?

Arrhenius eq:

K(T) = A e∧(-Ea/RT)
Ln K = Ln(A) - [(Ea/R)(1/T)]

∴ A: frecuency factor

∴ R = 8.314 E-3 KJ/K.mol

⇒ Ln K1 = Ln(A) - [Ea/R)*(1/T1)]..........(1)

⇒ Ln K2 = Ln(A) - [(Ea/R)*(1/T2)].............(2)

(1)/(2):

⇒ Ln (K1/K2) = (Ea/R)* (1/T2-1/T1)

⇒ Ln (K1/K2) = (116 KJ/mol/8.3134 E-3 KJ/K.mol)*(1/321.5 K - 1/298 K)

⇒ Ln (K1/K2) = (13952.37 K)*(- 2.453 E-4 K-1)

⇒ Ln (K1/K2) = - 3.422

⇒ K1/K2 = e∧(-3.422)

⇒ (3.32 E-10 s-1)/K2 = 0.0326

⇒ K2 = (3.32 E-10 s-1)/0.0326

⇒ K2 = 1.017 E-8 s-1

7 0

3 years ago