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Inessa [10]
2 years ago
14

Using Equation (10), calculate [Ag+] in the cell, where it is in equilibrium with 1 M Cl- ion. (Ecell in Equation (10) is the ne

gative of the measured value if the polarity is not the same as the standard cell.) Take [Cu2+] to be 1 M. Show your calculations. Ecell=0.4249v, Ecell=-0.00191. M E =E- 0.0592/2 • log ([Cu2+]/[Ag+1?) (10) cell
Chemistry
1 answer:
mixer [17]2 years ago
5 0

Answer:

7.16x10⁻⁸M = [Ag+]

Explanation:

Using the equation:

E(Cell) =E⁰ - 0.0592/2 • log ([Cu2+]/[Ag+]²)

<em>Where E</em>⁰<em>= 0.4249V</em>

<em>E(Cell) = -(-0.0019V) -Measured value-</em>

<em>[Cu2+] = 1M</em>

<em />

Replacing:

0.0019V = 0.4249V - 0.0592/2 • log (1M/[Ag+]²)

-0.423V = - 0.0296 • log (1M/[Ag+]²)

14.29 = log (1M/[Ag+]²)

1.95x10¹⁴ = 1M / [Ag+]²

[Ag+]² = 5.12x10⁻¹⁵M

7.16x10⁻⁸M = [Ag+]

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If a substance has a mass of 12.50 g and takes up 3.4 mL of space, what is the density?
miskamm [114]

Answer:

The answer is

<h2>3.68 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

<h3>density =  \frac{mass}{volume}</h3>

From the question

mass of substance = 12.50 g

volume = 3.4 mL

The density of the substance is

density =  \frac{12.50}{3.4}  \\  = 3.676470588...

We have the final answer as

<h3>3.68 g/mL</h3>

Hope this helps you

7 0
3 years ago
What is the chemical formula and net ionic equations for all three solutions.
timofeeve [1]

Answer:

See answer below

Explanation:

As you are asking for chemical formula and ionic equation, then, I will assume that after the station #4 below, are the solutions you are requiring.

You are also not specifing if you want for example, result of solution 1 + solution 3. If you need that, please post that on another question.

Now for the chemical formula, you need to identify the elements in all 3 solutions, and also the type of compound.

<u>1. Solution 2 Potassium Chloride: </u>

In this case we have Potassium on one side, and Chlorine on the other side, the symbol for those are K and Cl. As Potassium have the +1 oxidation state, cause is the only one that it can have, when it's next to an halide like chlorine or bromine, it will form a binary salt. The halides, usually work with the lowest oxydation state. In the case of Chlorine it will be -1, so, the formula will be:

KCl

And the net ionic equation will be the chemical equation that shows how the charges and atoms are balanced. In this case it would be:

K⁺ + Cl⁻ ------> KCl

<u>2. Solution 1 Copper(II) sulfate: </u>

In this case we have a tertiary salt, The copper's symbol is Cu, and is working with it oxydation state +2. Sulfate is an anion and it's formula is SO₄ and works with oxydation state -2 instead.

The chemical formula and ionic equation will be:  

Copper(II) sulfate: CuSO₄

And the net equation:

Copper sulfate: Cu²⁺ + SO₄²⁻ -------> CuSO₄

<u>3. Solution 3 Sodium hydroxide:</u>

In this case, we have a compound that it's usually used in acid base reactions. This is a strong base or hydroxide, and we have the element of Sodium (Na) with the oxydation state +1, is the only one it can have, and for the other side we have the oxydrile anion OH, and together is working with the oxydation state -1. So the chemical formula will be:

NaOH

And the net ionic equation:

Na⁺ + OH⁻ -------> NaOH

Hope this helps

4 0
3 years ago
How much heat in joules in needed to raise the temperature of 257g of ethanol c of ethanol =2.4j/gC by 49.1?
strojnjashka [21]

Answer:

30284.88J

Explanation:

c=mCtetha

c=257×2.4×49.1

hope it helps

please like and Mark as brainliest

6 0
3 years ago
How many cubic centimeters of an ore containing only 0.22% gold (by mass) must be processed to obtain $100.00 worth of gold? The
bezimeni [28]

Answer:

The cubic centimetres of the ore containing 0.22% gold (by mass) that must be processed to obtain the $100.00 worth of gold is approximately 216 cm³

Explanation:

The percentage by mass of gold in the ore = 0.22%

The density of the ore = 8.0 g/cm³

The price of the gold = $818 per troy ounce

14.6 troy oz = 1.0 pound

1 lb = 454 g

Given that one troy ounce = $818

$100 worth of gold = 1/818 ×100 troy ounce = 100/818 troy ounce

1 troy oz = 1.0/14.6 lb

100/818 troy oz =  100/818 × 1.0/14.6 lb = 250/29857 lb ≈ 0.0084 lb

1 lb = 454 g

250/29857 lb = 454 × 250/29857 g ≈ 3.8015 g

$100 = 3.8015 g worth of gold

The mass, M, of the ore containing 3.8015 g of gold is given as follows;

0.22% of M = 3.8015 g

0.22/100 × M = 3.8015 g

M = 3.8015 g × 100/0.22 = 1727.933 g

The volume, V, of the ore containing 3.8015 g of gold is given as follows;

Density of ore = Mass of ore/(Volume of ore)

Volume of ore = Mass of ore /(Density of ore)

The density of the ore = 8.0 g/cm³

Volume of ore = 1727.933 g /(8.0 g/cm³) = 215.99 cm³ ≈ 216 cm³

Therefore, the cubic centimetres of the ore containing 0.22% gold (by mass) that must be processed to obtain the $100.00 worth of gold ≈ 216 cm³.

5 0
3 years ago
What happens when a glass of water is set in the sun until the water
zalisa [80]

Answer:

D is the answer

Explanation:

6 0
2 years ago
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