CHEMISTRY COLLEGE

Determine the (H+), pH, and pOH of a solution with an [OH-] of 9.5 x 10-10 M at 25 °C. M pH =

Answers

Answer 1

Answer:

Answer : The concentration of $H^+$ ion, pH and pOH of solution is, $1.05* 10^(-5)M$ , 4.98 and 9.02 respectively.

Explanation : Given,

Concentration of $OH^-$ ion = $9.5* 10^(-10)M$

pH : It is defined as the negative logarithm of hydrogen ion or hydronium ion concentration.

The expression used for pH is:

$pH=-\log [H^+]$

First we have to calculate the pH.

$pOH=-\log [OH^-]$

$pOH=-\log (9.5* 10^(-10))$

$pOH=9.02$

The pOH of the solution is, 9.02

Now we have to calculate the pH.

$pH+pOH=14\n\npH=14-pOH\n\npH=14-9.02=4.98$

The pH of the solution is, 4.98

Now we have to calculate the $H^+$ concentration.

$pH=-\log [H^+]$

$4.98=-\log [H^+]$

$[H^+]=1.05* 10^(-5)M$

The $H^+$ concentration is, $1.05* 10^(-5)M$

Answer 2

Answer:

Answer:

pOH = 9.022, [H⁺] = 1.5×10⁻⁵ M, pH = 4.978

Explanation:

Given: [OH⁻] = 9.5 × 10⁻¹⁰ M, T= 25°C

As, pOH = - log [OH⁻]

⇒ pOH = - log (9.5 x 10⁻¹⁰) = 9.022

The self-ionisation constant of water is given by

Kw = [H⁺] [OH⁻] and pKw = pH + pOH

Since, at room temperature (25°C): Kw = 1.0 × 10⁻¹⁴ and pKw = 14.

Therefore, Kw = [H⁺] [OH⁻] = 1.0 × 10⁻¹⁴

⇒ [H⁺] = (1.0 × 10⁻¹⁴) ÷ [OH⁻] = (1.0 ×10⁻¹⁴) ÷ [9.5 × 10⁻¹⁰] = 0.105 ×10⁻⁴ = 1.5×10⁻⁵ M

also,

pH + pOH = pKw = 14

⇒ pH = 14 - pOH = 14 - 9.022 = 4.978

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Larry was told that a certain muscle cream was the newest best thing on the market and claims to double a person’s muscle power when used as part of a muscle-building workout. Interested in this product, he buys the special muscle cream and recruits Patrick and SpongeBob to help him with an experiment. Larry develops a special marshmallow weight-lifting program for Patrick and SpongeBob. He meets with them once every day for a period of 2 weeks and keeps track of their results. Before each session Patrick’s arms and back are lathered in the muscle cream, while Sponge Bob’s arms and back are lathered with the regular lotion.Which person is the control group?

SpongeBob

SpongeBob

Patrick

Patrick

Larry

Answers

SpongeBobs twin brother and Larry and Patrick’s twin brother

Final answer:

In the experiment, SpongeBob is the control group because regular lotion is used instead of the muscle cream. This allows a comparison with Patrick(rightly known as the experimental group) who uses the special muscle cream.

Explanation:

In the given scenario, SpongeBob represents the control group. In any experiment, the control group is the one that is kept normal or unchanged to be able to compare the effects of the variable being tested. In this case, it's the use of the special muscle cream. Patrick, whose arms and back are lathered with the muscle cream before each session, represents the experimental group because he is exposed to the variable being tested, which is the muscle cream. On the other hand, SpongeBob, who is given regular lotion instead of the special muscle cream, is part of the control group because he helps to provide a baseline for comparison.

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Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial chemist studying this reaction fills a 500. mL flask with 3.3 atm of sulfur dioxide gas and 0.79 atm of oxygen gas at 31.0 °C. He then raises the temperature, and when the mixture has come to equilibrium measures the partial pressure of sulfur trioxide gas to be 0.47 atm Calculate the pressure equilibrium constant for the reaction of sulfur dioxide and oxygen at the final temperature of the mixture. Round your answer to 2 significant digits.

Answers

Answer:

0.051

Explanation:

Let's consider the following reaction.

2 SO₂(g) + O₂(g) ⇄ 2 SO₃(g)

We can compute the pressures using an ICE chart.

2 SO₂(g) + O₂(g) ⇄ 2 SO₃(g)

I 3.3 0.79 0

C -2x -x +2x

E 3.3-2x 0.79-x 2x

The partial pressure of sulfur trioxide gas is 0.47 atm. Then,

2x = 0.47

x = 0.24

The pressures at equilibrium are:

pSO₂ = 3.3-2x = 3.3-2(0.24) = 2.82 atm

pO₂ = 0.79-x = 0.79-0.24 = 0.55 atm

pSO₃ = 0.47 atm

The pressure equilibrium constant (Kp) is:

Kp = pSO₃² / pSO₂² × pO₂

Kp = 0.47² / 2.82² × 0.55

Kp = 0.051

5.0 liters of a gas are at an initial pressure of 5.0 atmospheres. If the temperature and amount of a gas are kept constant, what is the new volume of the gas when pressure is increased to 7.0 atmospheres?

Answers

Final answer:

Using Boyle's Law of gases which states that the pressure and volume of a gas have an inverse relationship when temperature is kept constant, we find that when the pressure of the gas increases from 5.0 to 7.0 atmospheres, the volume of the gas decreases to approximately 3.57 liters.

Explanation:

The question pertains to the application of Boyle's Law, a fundamental concept in the field of physics dealing with gases. Boyle's Law states that the pressure and volume of a gas have an inverse relationship when the temperature is held constant. This means if the pressure of a gas increases, the volume decreases, and vice versa.

In this case, you have 5.0 liters of a gas under an initial pressure of 5.0 atmospheres. The pressure is then increased to 7.0 atmospheres, and you are asked to determine the new volume of the gas. To solve this problem, we use the formula for Boyle's Law, which is P1V1 = P2V2. We know P1 (initial pressure) is 5.0 atmospheres and V1 (initial volume) is 5.0 liters. P2 (final pressure) is increased to 7.0 atmospheres and V2 (final volume) is what we are trying to find.

So, we plug the numbers into the equation and get: 5.0 atmospheres * 5.0 liters = 7.0 atmospheres * V2. Solving for V2, we find V2 to be approximately 3.57 liters. Therefore, when the pressure of the gas is increased from 5.0 atmospheres to 7.0 atmospheres, the volume decreases to around 3.57 liters, while the temperature remains constant.

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Consider the following reaction: Pb(NO3)2(aq) + 2 NaCl(aq) → PbCl2(s) + 2 NaNO3(aq)If you react an excess of Pb(NO3)2with 26.3 g of NaCl, and you isolate 52.1 g of PbCl2, what is your percent yield?

Answers

Answer:

$\large \boxed{84.7 \, \%}$

Explanation:

Mᵣ: 58.44 278.11

Pb(NO₃)₂ + 2NaCl ⟶ PbCl₂ + 2NaNO₃

m/g: 26.3

1. Moles of NaCl

$\text{Moles of NaCl} = \text{26.3 g NaCl} * \frac{\text{1 mol NaCl}}{\text{58.44 g NaCl}} = \text{0.4505 mol NaCl}$

(b) Moles of PbCl₂

$\text{Moles of PbCl${_2}$} = \text{0.4505 mol NaCl} * \frac{\text{1 mol PbCl${_2}$}}{\text{2 mol NaCl}} = \text{0.2253 mol PbCl${_2}$}$

(c) Theoretical yield of PbCl₂

$\text{Mass of PbCl${_2}$} = \text{0.2253 mol PbCl${_2}$} * \frac{\text{278.11 g PbCl${_2}$}}{\text{1 mol PbCl${_2}$}} = \text{61.52 g PbCl${_2}$}$

(d) Percent yield

$\text{Percent yield} = \frac{\text{ actual yield}}{\text{ theoretical yield}} * 100 \,\% = \frac{\text{52.1 g}}{\text{61.52 g}} * 100 \, \% = \mathbf{84.7 \,\%}\n\n\text{The percent yield is $\large \boxed{\mathbf{84.7 \, \% }}$}$

A student conducts an experiment to see how music affects plant growth. The student obtains four identical plants. Each one is potted in the same type of soil and receives the sameamount of sunlight and water each day. Plant A listens to classical music for three hours each day. Plant B listens to rock music for three hours each day. Plant C listens to country
music for three hours each day. Plant D does not listen to any music at all.
1. Based on the experiment in the scenario, which visual aid would be most helpful in showing the change in the plants' heights over time?
O A. A line graph
O B. A pie chart
OC. A bar graph
O D. A timeline

Answers

Answer:

A. A line graph

Explanation:

You use line graphs to track changes over time. Line graphs are better when the changes are small. They are also more useful when you want to compare changes over the same period for more than one group, for example, plants exposed to music and a control group.

B is wrong. A pie chart is best for comparing parts of a whole.

C is wrong. You can use a bar graph to track changes over time, but small changes are harder to spot.

D is wrong. You use a timeline to mark important points in time, for example, when you are deciding the times when you must complete various stages of a project.

Which of the charts below do you think is more helpful in showing the change in plant height over time?

Aqueous solutions of sodium sulfide and copper(II) chloride are mixed together. Which statement is correct? a.Both NaCl and CuS will precipitate from solution b.No precipitate will form c.Only CuS will precipitate from solution d.Only NaCl will precipitate from solution