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miss Akunina [59]

2 years ago

15

Josh is studying pH calculations and tests a basic solution with a pH meter. He finds that the solution has a pH equal to 9.2. W

hat is the pOH of the solution? A) 2.5 B) 4.8 C) 9.6 D) 10.2

1 answer:

OleMash [197]2 years ago

5 0

POH is a measurement of the hydroxide concentration of a solution, and is calculated by 14 - pH. Therefore, your answer is B, 4.8.

You might be interested in

NH4NO3, whose heat of solution is 25.7 kJ/mol, is one substance that can be used in cold pack. If the goal is to decrease the te

makvit [3.9K]

Answer:

There are necessaries 35,2g of NH₄NO₃ per 100,0g of water to decrease the temperature of the solution from 25,0°C to 5,0°C

Explanation:

To decrease the temperature of the solution there are necessaries:

4,184J/g°C×(5,0°C-25,0°C)×(100,0g+X) = -Y

8368J + 83,68J/gX = Y <em>(1)</em>

Where x are grams of NH₄NO₃ you need to add and Y is the energy that you need to decrease the heat.

Also, the energy Y will be:

Y = 25700J/mol× $\frac{1mol}{80,043g}$ X

Y = 321J/g X <em>(2)</em>

Replacing (2) in (1)

8368J + 83,68J/g X = 321J/g X

8363J = 237,32J/gX

<em>X = 35,2g</em>

<em />

Thus, there are necessaries 35,2g of NH₄NO₃ per 100,0g of water to decrease the temperature of the solution from 25,0°C to 5,0°C

I hope it helps!

6 0

2 years ago

A solution is prepared by mixing 50.0 mL of 0.50 M Cu(NO3)2 with 50.0 mL of 0.50 M Co(NO3)2. Sodium hydroxide is added to the mi

vekshin1

Answer:

Cu(OH)₂ will precipitate first, with [OH⁻] = 2.97x10⁻¹⁰ M

Explanation:

The equilibriums that take place are:

Cu⁺² + 2OH⁻ ↔ Cu(OH)₂(s) ksp = 2.2x10⁻²⁰ = [Cu⁺²]*[OH⁻]²

Co⁺² + 2OH⁻ ↔ Co(OH)₂(s) ksp = 1.3x10⁻¹⁵ = [Co⁺²]*[OH⁻]²

Keep in mind that <em>the concentration of each ion is halved </em>because of the dilution when mixing the solutions.

For Cu⁺²:

2.2x10⁻²⁰ = [Cu⁺²]*[OH⁻]²

2.2x10⁻²⁰ = 0.25 M*[OH⁻]²

[OH⁻] = 2.97x10⁻¹⁰ M

For Co⁺²:

1.3x10⁻¹⁵ = [Co⁺²]*[OH⁻]²

1.3x10⁻¹⁵ = 0.25 M*[OH⁻]²

[OH⁻] = 7.21x10⁻⁸ M

<u>Because Copper requires less concentration of OH⁻ than Cobalt</u>, Cu(OH)₂ will precipitate first, with [OH⁻] = 2.97x10⁻¹⁰ M

3 0

2 years ago

Consider the following reactions and their respective equilibrium constants: NO(g)+12Br2(g)⇌NOBr(g)Kp=5.3 2NO(g)⇌N2(g)+O2(g)Kp=2

maxonik [38]

Answer:

Equilibrium constant of the given reaction is $1.3\times 10^{-29}$

Explanation:

$NO+\frac{1}{2}Br_{2}\rightleftharpoons NOBr$ .... $(K_{p})_{1}=5.3$

$2NO\rightleftharpoons N_{2}+O_{2}$ .... $(K_{p})_{2}=2.1\times 10^{30}$

The given reaction can be written as summation of the following reaction-

$2NO+Br_{2}\rightleftharpoons 2NOBr$

$N_{2}+O_{2}\rightleftharpoons 2NO$

......................................................................................

$N_{2}+O_{2}+Br_{2}\rightleftharpoons 2NOBr$

Equilibrium constant of this reaction is given as-

$\frac{[NOBr]^{2}}{[N_{2}][O_{2}][Br_{2}]}$

$=(\frac{[NOBr]}{[NO][Br_{2}]^{\frac{1}{2}}})^{2}(\frac{[NO]^{2}}{[N_{2}][O_{2}]})$

$=\frac{(K_{p})_{1}^{2}}{(K_{p})_{2}}$

$=\frac{(5.3)^{2}}{2.1\times 10^{30}}=1.3\times 10^{-29}$

7 0

2 years ago

Read 2 more answers

What are the possible products of the reduction of sulfuric acid? Select all that apply.

UNO [17]

Answer:

SO2(g) and H2S(g)

Explanation:

Sulphuric acid (H2SO4) is very important in industries and is often referred to as the best chemical due to its multiple uses. It is corrosive and used in making fertilizers and explosives. It can also be used to make paints, detergents etc.

Sulphuric acid reacts with Sodium iodide to give Hydrogen iodide and Sodium hydrogen Sulphate.

NaI (s) + H2SO4 (l) =HI (g) + NaHSO4 (s)

The Hydrogen iodide then reacts with the Sulphuric acid to give Iodine,Hydrogen sulphide and water

HI (g) + H2SO4 (l) = 4I2(s) + H2S (g) + 4 H2O (l)

The sulphuric acid can also undergo reduction reaction with hydrogen iodide to give Iodine, Sulphur dioxide and Water.

2 HI (g) + H2SO4 (l) = I2(s) + SO2 (g) + H2O (l)

7 0

2 years ago

The value of ka for nitrous acid (hno2) at 25 ∘c is 4.5×10−4. what is the value of δg at 25 ∘c when [h+] = 5.9×10−2m , [no2-] =

LuckyWell [14K]

To get the value of ΔG we need to get first the value of ΔG°:

when ΔG° = - R*T*㏑K

when R is constant in KJ = 0.00831 KJ

T is the temperature in Kelvin = 25+273 = 298 K

and K is the equilibrium constant = 4.5 x 10^-4

so by substitution:

∴ ΔG° = - 0.00831 * 298 K * ㏑4.5 x 10^-4

= -19 KJ

then, we can now get the value of ΔG when:

ΔG = ΔG° - RT*㏑[HNO2]/[H+][NO2]

when ΔG° = -19 KJ

and R is constant in KJ = 0.00831

and T is the temperature in Kelvin = 298 K

and [HNO2] = 0.21 m & [H+] = 5.9 x 10^-2 & [NO2-] = 6.3 x 10^-4 m

so, by substitution:

ΔG = -19 KJ - 0.00831 * 298K* ㏑(0.21/5.9x10^-2*6.3 x10^-4 )

= -40

8 0

2 years ago