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What is the pH of a 0.45 M solution of hydrochloride acid (H

Answers

Answer 1

Answer:

Answer:

0.35

Explanation:

pH = -log[H+]

[H+] = [HCl} = 0.45 M because HCl is a strong acid, and dissociate completely.

pH = - log[0.45] = 0.35

Related Questions

Refer to the following standard reduction half-cell potentials at 25∘C: VO2+(aq)+Ni2+(aq)2H+(aq)++2e−e−→ →Ni(s)VO2+(aq) +H2O(l)E∘=−0.23V E∘=0.99V An electrochemical cell is based on these two half-reactions: Oxidation:Reduction:Ni(s)VO2+(aq,0.083M)+2H+(aq,1.1M)+e−→→Ni2+(aq,2.5M)+2e−VO2+(aq,2.5M)+H2O(l) Calculate the cell potential under these nonstandard concentrations. Express the cell potential to two decimal places and include the appropriate units.
g Acetic acid is diluted with water to make a solution of vinegar. You have a sample of vinegar that contains 16.7 g of acetic acid. Determine the number of moles of acetic acid in the vinegar sample.
I have always enjoyed eating tuna fish. Unfortunately,a study of the mercury content of canned tuna in 2010 foundthat chunk white tuna contains 0.6 ppm Hg and chunk lighttuna contains 0.14 ppm. (S. L. Gerstenberger, A. Martinson,and J. L. Kramer, Environ. Toxicol. Chem. 2010, 29, 237.) TheU.S. Environmental Protection Agency recommends no morethan 0.1 mg Hg/kg body weight per day. I weigh 68 kg. Howoften may I eat a can containing 6 ounces (1 lb 5 16 oz) ofchunk white tuna so that I do not average more than 0.1 mgHg/kg body weight per day? If I switch to chunk light tuna,how often may I eat one can?
a chemist encounters an unknown metal. They drop the metal into a graduated cylinder containing water, and find the volume change is 3.2 mL. If the metal weighs 1.5g, what is the density in g/mL of the metal?
Juan noticed that a door that sticks on hot days does not stick when it cools off at night. This is because the metal contracts as it cools. The reason for this is that, as the metal cools, the atoms:become smaller and take up less space .slow down and move closer together.merge together to form fewer atoms.speed up and move further apart.

As the speed of the particles decreases, -a. Intermolecular forces become stronger

b. Intermolecular forces become weaker

c. Intermolecular forces do not change

d. Energy increases
Engshus
ratoway

Answers

As the speed of the particles decreases, intermolecular forces become stronger. Thus, the correct option for this question is A.

What are Intermolecular forces?

The intermolecular forces may be defined as the forces of attraction. that is present between atoms, molecules, and ions when they are placed close to each other in order to form a compound or element. This force is continuously acting on the neighboring particles of different molecules.

It is found that at low temperature, when the speed of molecules/particles decrease, they migrate closer to one another that results in their intermolecular forces that become stronger as compared to the initial one.

As the attraction between molecules gradually increases, their movement decreases, and undergo fewer collisions between them.

Therefore, as the speed of the particles decreases, intermolecular forces become stronger. Thus, the correct option for this question is A.

To learn more about Intermolecular forces, refer to the link:

brainly.com/question/13588164

#SPJ2

Answer:

Explanation:

Calculate the solubilities of the following compounds in a 0.02 M solution of barium nitrate using molar concentrations, first ignoring ionic strength and activities. a. silver iodate
b. barium sulfate
c. Repeat the above calculations using ionic strength and activities.

Answers

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)$$^(-)$]} = s* s = s^(2) = 3.0* 10^(-8)\ns = \sqrt{3.0* 10^(-8)} \text{ mol/L} = 1.7 * 10^(-4) \text{ mol/L}$

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-)$]} = (0.02 + s) * s \approx 0.02s = 1.1* 10^(-10)\ns = (1.1* 10^(-10))/(0.02) \text{ mol/L} = 5.5 * 10^(-9) \text{ mol/L}$

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

$\mu = (1)/(2) \sum_(i) {c_(i)z_(i)^(2)}\n\n\mu = (1)/(2) (\text{[Ba$^(2+)$]}\cdot (2+)^(2) + \text{[NO$_(3)$$^(-)$]}*(-1)^(2)) = (1)/(2) (\text{0.02}* 4 + \text{0.04}*1)= (1)/(2) (0.08 + 0.04)\n\n= (1)/(2) *0.12 = 0.06$

(ii) Silver iodate

a. Calculate the activity coefficients of the ions

$\log \gamma = -0.51z^(2)√(I) = -0.051(1)^(2)√(0.06) = -0.51* 0.24 = -0.12\n\gamma = 10^(-0.12) = 0.75$

b. Calculate the solubility

AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq)

$K_(sp) =\text{[Ag$^(+)$]$\gamma_(Ag^(+))$[IO$_(3)$$^(-)$]$\gamma_{IO_(3)^(-)}$} = s*0.75* s * 0.75 =0.56s^(2)= 3.0 * 10^(-8)\ns^(2) = (3.0 * 10^(-8))/(0.56) = 5.3 * 10^(-8)\n\ns =2.3 * 10^(-4)\text{ mol/L}$

(iii) Barium sulfate

a. Calculate the activity coefficients of the ions

$\log \gamma = -0.51z^(2)√(I) = -0.051(2)^(2)√(0.06) = -0.51*16* 0.24 = -0.50\n\gamma = 10^(-0.50) = 0.32$

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-)}$} = (0.02 + s) * 0.32* s* 0.32 \approx 0.02*0.10s\n2.0* 10^(-3)s = 1.1 * 10^(-10)\ns = (1.1* 10^(-10))/(2.0 * 10^(-3)) \text{ mol/L} = 5.5 * 10^(-8) \text{ mol/L}$

Pharmacists sometimes measure medicines in the unit of "grains," where 1 gr = 65 mg. The label on a bottle of aspirin reads "Aspirin, 5 gr." How many mg of aspirin are in a tablet containing 5 gr?

Answers

Answer:

325mg of Aspririn

Explanation:

First you should note the information that the problem gives you:

- The bottle of Aspirin has 5gr (grains)

- 1gr(grain) = 65mg (miligrams)

Also, the problem is asking about how many aspirin are in 5 gr (grains), so you should use a conversion factor, as follows:

-First you should put the quantity you need to convert:

$5grAspirin$

-Then you write the denominator of the conversion factor that must have the same units that you want to convert, in this case gr:

$5grAspirin*\frac{}{1grAspirin}$

-Then you write the numerator with the units that you want to obtain and the numerical equivalence between the units, in this case:

$5grAspirin*(65mgAspririn)/(1grAspirin)$

-Finally you multiply numerators and divide by denominators:

$5grAspirin*(65mgAspririn)/(1grAspirin)=325mgAspririn$

Be sure to answer all parts. Consider both 5-methyl-1,3-cyclopentadiene (A) and 7-methyl-1,3,5-cycloheptatriene (B). Which labeled H atom is most acidic? Hb is most acidic because its conjugate base is aromatic. Hc is most acidic because its conjugate base is antiaromatic. Ha is most acidic because its conjugate base is antiaromatic. Hd is most acidic because its conjugate base is aromatic. Which labeled H atom is least acidic? Ha is least acidic because its conjugate base is aromatic. Hb is least acidic because its conjugate base is antiaromatic. Hd is least acidic because its conjugate base is aromatic. Hc is least acidic because its conjugate base is antiaromatic.

Answers

Due to the conjugate base of the hydrogen atom is aromatic, Hb is regarded as the most acidic. Because the conjugate base of the hydrogen atom Hc is anti-aromatic, it is the least acidic.

The correct options are:

(A) - (a)

(B) - (d)

What are the most and the least acidic hydrogen atom?

The hydrogen connected at the heptatriene's tertiary position (at the 7-methyl) would be particularly acidic, as its removal would leave a positive charge that could be transported around the ring via resonance.

The hydrogen connected to the pentadiene (5-methyl) at the tertiary position would not be acidic, as removing it would result in an anti-aromatic structure.

Thus, the least acidic H atom is Hc and the most acidic H atom is Hb.

Learn more about hydrogen atom, here:

brainly.com/question/7916557

I don’t have a picture but I can describe it to you.

The hydrogen that is attached at the tertiary position on the heptatriene (at the 7-methyl) would be very acidic, as removal would leave a positive charge that could be moved throughout the ring through resonance. This would mean that the three double bonds would be participating in resonance, and the deprotonated structure would be aromatic, thus making this favorable.

The hydrogen that is attached at the tertiary position on the pentadiene (5-methyl) would NOT be acidic, as removal would cause an antiaromatic structure.

Any other hydrogens would NOT be acidic. Those vinylic to their respective double bonds would seriously destabilize the double bond if removed, and hydrogens attached to the methyl group jutting off the ring have no incentive to leave the carbon.

Hope this helps!

The change in velocity over a specific amount of time Includes speeding up. slowing down or changing direction. This is known as O speed acceleration changing speed

Answers

Answer:

Acceleration

Explanation:

Acceleration = $(velocity)/(time)$

When the reaction mixture is worked-up, it is first washed three times with 5% sodium bicarbonate, and then with a saturated nacl solution. explain why?

Answers

Solution:

After the reaction of mixture is worked-up Washing three times the organic with sodium carbonate helps to decrease the solubility of the organic layer into the aqueous layer. This allows the organic layer to be separated more easily.

And then the reaction washed by saturated NACL we have The bulk of the water can often be removed by shaking or "washing" the organic layer with saturated aqueous sodium chloride (otherwise known as brine). The salt water works to pull the water from the organic layer to the water layer.

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