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2 years ago

Determine the [H+], [OH-], pH, and pOH for 0.045 M HBr

Chemistry

1 answer:

elena-s [515]2 years ago

5 0

Answer:

The answer to your question is below

Explanation:

Data

[HBr] = 0.045 M

[H⁺] = ?

[OH⁻] = ?

pH = ?

pOH = ?

Process

1.- Determine the [H⁺]

The [H⁺] is equal to the [HBr] then, [H⁺] = 0.045 M

2.- Determine the pH

pH = -log[H⁺]

-Substitution

pH = -log[0.045]

-Result

pH = 1.35

3.- Determine the pOH

Formula

pH + pOH = 14

-Solve for pOH

pOH = 14 - pH

-Substitution

pOH = 14 - 1.35

-Result

pOH = 12.65

4.- Determine the [OH⁻]

pOH = -log[OH⁻]

-Solve for [OH⁻]

[OH⁻] = antilog(-pOH)

-Substitution

[OH⁻] = antilog (-12.65)

-Result

[OH⁻] = 2.22 x 10⁻¹³

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In a 1 M solution of NH3(aq), identify the relative molar amounts of these species. Arrange them from most to least.

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Answer:

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2 years ago

A solution contains 10.0 g pentane, C5H12, 20.0 g hexane, C6H14, and 10.0 g benzene, C6H6. What is the mole fraction of hexane?

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Answer:

b) 0.47

Explanation:

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2 years ago

You've just solved a problem and the answer is the mass of an electron, me=9.11×10−31kilograms. How would you enter this number

irina1246 [14]

Answer: $9.11\times 10^{-31}kg$

Explanation:

Significant figures : The figures in a number which express the value or the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

Rules for significant figures:

Digits from 1 to 9 are always significant and have infinite number of significant figures.

All non-zero numbers are always significant.

All zero’s between integers are always significant.

All zero’s after the decimal point are always significant.

All zero’s preceding the first integers are never significant.

Thus $9.11\times 10^{-31}kg$ has three significant figures

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2 years ago

For the reaction A + B − ⇀ ↽ − C + D A+B↽−−⇀C+D , assume that the standard change in free energy has a positive value. Changing

AURORKA [14]

Answer:

a. Not change the free energy value

b. Increase the free energy value

c. Decrease the free energy value

d. Decrease the free energy value

Explanation:

a. Adding a catalyst:

A catalyst is a substance that will reduce the activation energy of a reaction, it means that the reaction will occur fast. The values of enthalpy, entropy, and free energy are not affected by a catalyst, so ΔG remains the same.

b. Increasing [C] and [D]:

For a reversible reaction, the value of free energy can be calculated by:

ΔG = ΔG° + RT*lnK

Where ΔG° is the standard value for free energy, R is the gas constant, T is the temperature, and K is the constant of equilibrium, which in this case:

K = ([C]*[D])/([A]*[B])

When [C] and [D] increase, the value of K increases, and lnK also increases, then, the value of ΔG increases.

c. Coupling with ATP hydrolysis:

The free energy can be calculated by:

ΔG = ΔH - TΔS

Where ΔH is the change in enthalpy, and ΔS the change in entropy. The ATP hydrolysis is an exothermic reaction, so ΔH <0. When it is coupled, it will reduce the total value of ΔH, and because of that, the value of ΔG will decrease.

d. Increasing [A] and [B]:

As explained above, the increasing at [A] and [B] will decrease the value of K, so the value of lnK will decrease, and ΔG value will also decrease.

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1 year ago

Read 2 more answers

Carbon dioxide readily absorbs radiation with an energy of 4.67 x 10-20 J. What is the wavelength and frequency of this radiatio

Tanya [424]

Answer:

ν = 7.04 × 10¹³ s⁻¹

λ = 426 nm

It falls in the visible range

Explanation:

The relation between the energy of the radiation and its frequency is given by Planck-Einstein equation:

E = h × ν

where,

E is the energy

h is the Planck constant (6.63 × 10⁻³⁴ J.s)

ν is the frequency

Then, we can find frequency,

$\nu = \frac{E}{h}= \frac{4.67 \times 10^{-20}J }{6.63 \times 10^{-34}J.s} = 7.04 \times 10^{13} s^{-1}$

Frequency and wavelength are related through the following equation:

c = λ × ν

where,

c is the speed of light (3.00 × 10⁸ m/s)

λ is the wavelength

$\lambda = \frac{c}{\nu } =\frac{3.00 \times 10^{8} m/s }{7.04 \times 10^{13} s^{-1} } =4.26 \times 10^{-6}m.\frac{10^{9}nm }{1m} = 426 nm$

A 426 nm wavelength falls in the visible range (≈380-740 nm)

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