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Y_Kistochka [10]
2 years ago
7

When heated, KClO3 decomposes into KCl and O2. 2KClO3⟶2KCl+3O2 If this reaction produced 13.2 g KCl, how many grams of O2 were p

roduced?
Chemistry
1 answer:
MissTica2 years ago
7 0

Answer:

8.5gm O2 produced

Explanation:

When heated, KClO3 decomposes into KCl and O2. 2KClO3⟶2KCl+3O2 If this reaction produced 13.2 g KCl, how many grams of O2 were produced?

for every 2 moles of KCl produced, 3 moles of O2 are produced

Mol weight of KCl =39+35.5=74.5gm

13.2 gm KCl = 13.2/74.5 = 0.177 moles

this will make (3/2) X 0.177 = 0.2655 moles of O2

O2 mol wt is 32  0.2655 X32 = 8.5gm O2 produced

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sladkih [1.3K]

Answer:

[α] = -77.5° / \frac{\textup{dm-g}}{\textup{mL}}

Explanation:

Given;

Mass of optically pure substance in the solution = 10 g

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measured rotation = - 3.10°

Now,

The specific rotation ( [α] ) is given as:

[α] = \frac{\alpha}{c\times l}

here,

α is the measured rotation = -3.10°

c is the concentration

or

c = \frac{\textup{Mass of optically pure substance in the solution}}{\textup{Volume of water}}

or

c =  \frac{10}{500}

or

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7 0
3 years ago
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6 0
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beks73 [17]

Answer:

6.4 L

Explanation:

When all other variables are held constant, you can use Boyle's Law to find the missing volume:

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In this equation, "P₁" and "V₁" represent the initial pressure and volume. "P₂" and "V₂" represent the final pressure and volume. You can find the theoretical volume by plugging the given values into the equation and simplifying.

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6.4 = (1.0 atm)V₂                                           <----- Simplify left side

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6 0
2 years ago
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Wewaii [24]

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Why did I multiply the mass in this equation by 2? Am I pulling your leg?

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Rearranging it [OH-] = 10^-pH

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