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Mandarinka [93]

3 years ago

8

How many liters of oxygen gas, at STP, are produced by the decomposition of 0.605 moles of potassium chlorate? 2KClO3 ➞ 2KCl + 3

O2

1 answer:

grigory [225]3 years ago

6 0

Answer:

19 dollar fortnite card who wants it

Explanation:

and yes it is free

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What tests can you do to identify gases, based on their chemical properties?

GuDViN [60]

Tests for gases
Hydrogen, oxygen, carbon dioxide, ammonia and chlorine can be identified using different tests.
Hydrogen. A lighted wooden splint makes a popping sound in a test tube of hydrogen.
Oxygen. A glowing wooden splint relights in a test tube of oxygen.

5 0

3 years ago

Bezzdna [24]

A. I think sorry if it’s wrong

5 0

3 years ago

Please help!! Why is the following Electron Configuration incorrect for Aluminium?

Margarita [4]

Answer:

3s^1 would be 3s^2

Explanation:

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3 years ago

What has to be true of a substance for it to dissolve in water?

lord [1]

The polarity of a water molecule comes from the uneven distribution of electron density of hydrogen and oxygen atom. The oxygen in the water molecule is more electronegative than the hydrogen. Water has a partial positive charge near the hydrogen atom and a partial negative charge near the oxygen atom. The result of this electrostatic attraction results in the bond called hydrogen bond. Also, because of this bond, it has the ability to dissolve most of the solutes due to its polarity and bonding.

7 0

3 years ago

Hexane and octane are mixed to form a 45 mol% hexane solution at 25 deg C. The densities of hexane and octane are 0.655 g/cm3 an

avanturin [10]

Answer:

The required volume of hexane is 0.66245 Liters.

Explanation:

Volume of octane = v=1.0 L= $1000 cm^3$

Density of octane= d = $0.703 g/cm^3$

Mass of octane ,m= $d\times v=0.703 g/cm^3\times 1000 cm^3=703 g$

Moles of octane = $\frac{m}{114 g/mol}=\frac{703 g}{114 g/mol}=6.166 mol$

Mole percentage of Hexane = 45%

Mole percentage of octane = 100% - 45% = 55%

$55\%=\frac{6.166 mol}{\text{Total moles}}\times 100$

Total moles = 11.212 mol

Moles of hexane :

$45%=\frac{\text{moles of hexane }}{\text{Total moles}}\times 100$

Moles of hexane = 5.0454 mol

Mass of 5.0454 moles of hexane,M = 5.0454 mol × 86 g/mol=433.9044 g

Density of the hexane,D = $0.655 g/cm^3$

Volume of hexane = V

$V=\frac{M}{D}=\frac{433.9044 g}{0.655 g/cm^3}=662.4494 cm^3\approx 0.66245 L$

(1 cm^3= 0.001 L)

The required volume of hexane is 0.66245 Liters.

5 0

3 years ago