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alexandr402 [8]
3 years ago
7

3. Determine the pH of each of the following solutions. (Hint: See Sample Problem B.) a. 1.0 x 10-2 M HCI c. 1.0 x 10-MHI b. 1.0

x 10-3 M HNO3 d. 1.0 x 10-M HB​
Chemistry
1 answer:
densk [106]3 years ago
3 0

Answer:

a. pH = 2 b. pH = 3 c. pH = 1 d. Unanswerable

Explanation:

pH = -log[H+] OR pH = -log{H3O+]

                    and inversely

pOH = -log[OH-]

1. Determine what substance you are working with, (acid/base)

2. Determine whether or not that acid or base is strong or weak.

a. 1.0 x 10^-2M HCl

HCl is a strong acid, therefore it will dissociate completely into H+ and Cl- with all ions going to the H+, therefore, the concentration of HCl and concentration of H+ are going to be equal, meaning we simply take the negative logarithm of the concentration of HCl and that would equal pH

pH = -log[H+]

pH = -log(1.0x10^-2)

pH = 2

b. 1.0 x 10^-3M HNO3

HNO3 like part a, is a strong acid, therefore it would simply require you to take the negative logarithm of the concentration of the compound itself, to find its pH.

pH = -log[H+]

pH = -log(1.0 x 10^-3)

pH = 3

c. 1.0 x 10^-1M HI

Like the previous parts, HI is a strong acid

pH = -log[H+]

pH = -log(0.10)

pH = 1

d. HB isn't an element, nor is it a compound so that would be unanswerable.

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Answer:

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8 0
3 years ago
Which of the following is NOT a reason that carbon is the basis of organic chemistry? Question 3 options:
mina [271]
I believe the best reason as for why carbon is the basis of organic chemistry would be D. Carbon can bond with hydrogen.
5 0
4 years ago
If 29.4 mL of ethanol is dissolved in water to make 359 mL of solution, what is the concentration expressed in volume/volume % o
Sveta_85 [38]

Answer: 8.2\%

Explanation:- Volume percentage is the ratio of volume of solute to the volume of solution defined in terms of percentage.

{\text {volume percentage}}=\frac{\text {volume of solute}}{\text {volume of solution}}\times 100\%

Given: volume of solute = 29.4 ml

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{\text {volume percentage}=\frac{29.4}{359}\times 100\%=8.2\%


6 0
4 years ago
The equilibrium constant is equal to 5.00 at 1300 K for the reaction:2 SO2(g) + O2(g) ⇌ 2 SO3(g). If initial concentrations are
oee [108]

This is an incomplete question, here is a complete question.

The equilibrium constant is equal to 5.00 at 1300 K for the reaction:

2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)

If initial concentrations are [SO₂] = 1.20 M, [O₂] = 0.45 M, and [SO₃] = 1.80 M, the system is

A) at equilibrium.

B) not at equilibrium and will remain in an unequilibrated state.

C) not at equilibrium and will shift to the left to achieve an equilibrium state.

D) not at equilibrium and will shift to the right to achieve an equilibrium state.

Answer : The correct option is, (A) at equilibrium.

Explanation :

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given balanced chemical reaction is,

2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)

The expression for reaction quotient will be :

Q=\frac{[SO_3]^2}{[SO_2]^2[O_2]}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

Q=\frac{(1.80)^2}{(1.20)^2\times (0.45)}=5.0

The given equilibrium constant value is, K_c=5.00

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

There are 3 conditions:

When Q>K_c that means product > reactant. So, the reaction is reactant favored.

When Q that means reactant > product. So, the reaction is product favored.

When Q=K_c that means product = reactant. So, the reaction is in equilibrium.

From the above we conclude that, the Q=K_c that means product = reactant. So, the reaction is in equilibrium.

Hence, the correct option is, (A) at equilibrium.

7 0
3 years ago
Why the standard enthalpy change of hydration of copper sulphate cannot be measured directly ?
Juliette [100K]

It is difficult to measure the enthalpy change of hydration accurately in a direct way because the hydration process can't be controlled directly. Instead, anhydrous and hydrated copper(II) sulfate can be dissolved in water. Each 'route' produces a solution of hydrated copper(II) sulfate.

5 0
3 years ago
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