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2 years ago

What is the oxidation state of an individual bromine atom in kbro2?

1 answer:

6 0

The oxidation state of potassium ion K = +1
The oxidation state of oxygen ion O = -2
So, the oxidation state of O2 is = -2 x 2 = -4
Since, KBrO2 is neutral so,
(+1) + (x) + (-4) = Zero
-3 + X = Zero
So, X = +3
The oxidation state of individual bromine atom in KBrO2 is +3

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A flask contains 0.180 mol of liquid bromine, br2. determine the number of bromine molecules present in the flask.

Reika [66]

Avogadro's number represents the number of units in one mole of any substance. This has the value of 6.022 x 10^23 units / mole. This number can be used to convert the number of atoms or molecules into number of moles. We calculate as follows:

0.180 mol Br2 ( 6.022 x 10^23 molecules / mole ) = 1.084x10^23 molecules Br2

5 0

2 years ago

Read 2 more answers

In a chemistry lab, you have two vinegars. one is 5% acetic acid, and one is 6.5% acetic acid. you want to make 200 ml of a vine

nexus9112 [7]

Assume that the amount needed from the 5% acid is x and that the amount needed from the 6.5% acid is y.

We are given that:
The volume of the final solution is 200 ml
This means that:
x + y = 200
This can be rewritten as:
x = 200 - y .......> equation I

We are also given that:
The concentration of the final solution is 6%
This means that:
5%x + 6.5%y = 6% (x+y)
This can be rewritten as:
0.05 x + 0.065 y = 0.06 (x+y) ............> equation II

Substitute with equation I in equation II and solve for y as follows:
0.05 x + 0.065 y = 0.06 (x+y)
0.05 (200-y) + 0.065 y = 0.06 (200-y+y)
10 - 0.05 y + 0.065 y = 12
0.015y = 12-10 = 2
y = 2/0.015
y = 133.3334 ml

Substitute with the y in equation I to get the x as follows:
x = 200 - y
x = 200 - 133.3334
x = 66.6667 ml

Based on the above calculations:
The amount required from the 5% acid = x = 66.6667 ml
The amount required from the 6.5% acid = y = 133.3334 ml

Hope this helps :)

6 0

2 years ago

Read 2 more answers

In the activity, click on the E∘cell and Keq quantities to observe how they are related. Use this relation to calculate Keq for

joja [24]

Answer:

ΔG° = -118x10³ J/mol

Explanation:

The two half-reactions in the cell are:

Oxidation half-reaction:

Co(s) → Co²⁺(aq) + 2e⁻; E° = -0,28V

Reduction half-reaction:

Cu²⁺(aq)+2e⁻ → Cu(s); E° = 0,34V

The E° of the cell is defined as:

$E_{cell} = E_{red} - E_{ox}$

Replacing:

0,34V - (-0,28V) = 0,62V

It is possible to obtain the keq from E°cell with Nernst equation thus:

nE°cell/0,0592 = log (keq)

Where:

E°cell is standard electrode potential (0.62 V)

n is number of electrons transferred (2 electrons, from the half-reactions)

Replacing:

0,62V×2/0,0592 = log (keq)

20,946 = log keq

keq = 8,83x10²⁰≈ 5,88x10²⁰

ΔG° is defined as:

ΔG° = -RT ln Keq

Where R is gas constant (8,314472 J/molK) and T is temperature (298K):

ΔG° = -8,314472 J/molK×298K ln5,88x10²⁰

ΔG° = -118x10³ J/mol

I hope it helps!

8 0

2 years ago

According to the law of conservation of mass, if an element A has an atomic mass of 2 mass units and element B has an atomic mas

LenaWriter [7]

The law of conservation of mass states that mass is neither created nor destroyed. Since we have 2 g/mol of A and 3 g/mol of B then AB should be equal to the sum of their molar mass that is

2 g/mol + 3 g/mol = 5 g/mol AB

for the case of A2B3

A2 = 2 * 2 = 4 g/mol
B3 = 3 * 3 = 9 g/mol

therefore A2B3 = 13 g/mol

5 0

2 years ago

A mixture of three gases has a total pressure at 298 K of 1560 mm Hg. the mixture is analyzed and is found to contain 1.50 mol N

Agata [3.3K]

Answer:

the partial pressure of Xe is 452.4 mmHg

Explanation:

Dalton's law of partial pressures says that in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of the individual gases.

The partial pressures can be calculated with the molar fraction of the gas, in this case, Xe.

Molar fraction of Xe is calculated as follows:

$x_{Xe}=\frac{n_{Xe} }{n_{t} }$

$x_{Xe}=1.75/5.9\\x_{Xe}=0.29$

Then, 0.29 is the molar fraction of Xe in the mixture of gases given.

To know the parcial pressure of Xe, we have to multiply the molar fraction by the total pressure:

Partial Pressure of Xe=1560mmHg*0.29

Partial Pressure of Xe=452.4mmHg

7 0

1 year ago