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3 years ago

How many hydrogen atoms are in 5.40 mol of ammonium sulfide?

1 answer:

3 0

Answer Expert Verified
Assuming that the ammonium sulfide formula is (NH4)2S then you can see that there are 2 nitrogen, 8 hydrogen and 2 sulfur atoms for every ammonium sulfide.
Hope that helps

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Given that a for HCN is 6. 2×10^−10 at 25 °C. What is the value of b for cn− at 25 °C?

Kay [80]

If Ka for HCN is 6. 2×10^−10 at 25 °C, then the value of Kb for cn− at 25 °C is 1.6 × 10^(-5).

<h3>What is base dissociation constant? </h3><h3 />

The base dissociation constant (Kb) is defined as the measurement of the ions which base can dissociate or dissolve in the aqueous solution. The greater the value of base dissociation constant greater will be its basicity an strength.

The dissociation reaction of hydrogen cyanide can be given as

HCN --- (H+) + (CN-)

Given,

The value of Ka for HCN is 6.2× 10^(-10)

The correlation between base dissociation constant and acid dissociation constant is

Kw = Ka × Kb

Kw = 10^(-14)

Substituting values of Ka and Kw,

Kb = 10^(-14) /{6.2×10^(-10) }

= 1.6× 10^(-5)

Thus, the value of base dissociation constant at 25°C is 1.6 × 10^(-5).

learn more about base dissociation constant :

brainly.com/question/9234362

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4 0

2 years ago

Heat and pressure deep in the earth change shale to slate. Why does this happen?

dedylja [7]

Answer:

Because of the ground above and the molten iron core.

Explanation:

As you go deeper into the earth you have more and more of the earth above you that applies pressure on you. The heat near the very top of the plate is relatively low and it decreases by a little as you go down. But once you start to go down into the earth by hundreds of meters then the temperature starts to rise and it rises pretty quickly due to the heat being transferred from the molten iron outer core of the earth.

5 0

3 years ago

One example of a physical change is

Gekata [30.6K]

Plucking eyebrows hehe

8 0

3 years ago

Read 2 more answers

Use the Nernst equation to calculate the concentration of the unknown solution. Base this on your experimental voltage of 1.074

Hoochie [10]

Answer:

0.3793 M

Explanation:

The unknown metal is zinc. So the equation of the reaction is;

Zn(s) + Cu^2+(aq) -------> Zn^2+(aq) + Cu(s)

From Nernst equation;

E = E° - 0.0592/n log Q

[Cu2+] = 0.050179 M

n = 2

[Zn^2+] = ?

E = 1.074 V

E° = 0.34 - (-0.76) = 1.1 V

Substituting values;

1.074 = 1.1 - 0.0592/2 log [Zn^2+]/0.050179

1.074 - 1.1 = - 0.0592/2 log [Zn^2+]/0.050179

-0.026 = -0.0296 log [Zn^2+]/0.050179

-0.026/-0.0296 = log [Zn^2+]/0.050179

0.8784 =log [Zn^2+]/0.050179

Antilog(0.8784) = [Zn^2+]/0.050179

7.558 = [Zn^2+]/0.050179

[Zn^2+] = 7.558 * 0.050179

[Zn^2+] = 0.3793 M

4 0

3 years ago

What is the freezing point of a solution of 498mL of water (solute) dissolved in 2.50 L of ethanol (solvent), C2H5OH? The densit

jok3333 [9.3K]

Answer:

Freezing T° of solution is -142.4°C

Explanation:

This excersise is about colligative properties, in this case freezing point depression,

ΔT = Kf . m . i

Where ΔT = Freezing T° of solvent - Freezing T° of solution

Kf = Cryoscopic constant

m = mol/kg (molality)

i = Number of ions dissolved.

Water is not ionic, so i = 1

Let's find out m.

We determine mass of water, by density

498ml . 1 g/mL = 498 g

We convert the mass of water to moles → 498 g . 1mol/18g = 27.6 moles

We determine mass of solvent by density

2500 mL . 0.789 g/mL = 1972.5 g

Notice, we had to convert L to mL to cancel units.

1 cm³ = 1 mL

We convert the mass from g to kg → 1972.5 g . 1kg /1000g = 1.97kg

We determine m = mol/kg → 27.6mol / 1.97kg = 13.9 m

Kf for ethanol is: 1.99 °C/m

Freezing T° for ethanol is: -114.6°C

We replace at formula: - 114.6°C - Freezing T° solution = 1.99 °C/m . 13.9 m . 1

- 114.6°C - Freezing T° solution = 27.8 °C

- Freezing T° solution = 27.8°C + 114.6°C

Freezing T° Solution = - 142.4 °C

7 0

2 years ago