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3 years ago

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Chemistry

1 answer:

Pani-rosa [81]3 years ago

5 0

Answer : The mass of of water present in the jar is, 298.79 g

Solution : Given,

Mass of barium nitrate = 27 g

The solubility of barium nitrate at $20^oC$ is 9.02 gram per 100 ml of water.

As, 9.02 gram of barium nitrate present in 100 ml of water

So, 27 gram of barium nitrate present in $\frac{27g}{9.02g}\times 100ml=299.33ml$ of water

The volume of water is 299.33 ml.

As we know that the density of water at $20^oC$ is 0.9982 g/ml

Now we have to calculate the mass of water.

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}$

$\text{Mass of water}=(0.9982g/ml)\times (299.33ml)=298.79g$

Therefore, the mass of of water present in the jar is, 298.79 g

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For the reaction shown, identify the element oxidized, the element reduced, the oxidizing agent, and the reducing agent. KNO3 →

Vinvika [58]

Answer : The oxidizing element is N and reducing element is O.

$KNO_{3}$ is act as an oxidizing agent as well as reducing agent.

Explanation :

An Oxidizing agent is the agent which has ability to oxidize other or a higher in oxidation number.

Reducing agent is the agent which has ability to reduce other or lower in oxidation number.

The given reaction is :

$KNO_{3} \rightarrow KNO_{2} +O_{2}$

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The oxidation number of N in $KNO_{3}$ is calculated as:

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$KNO_{3}$ act as a reducing agent.

$KNO_{3} \rightarrow KNO_{2} +O_{2}$

The oxidation number of O in $KNO_{3}$ is calculated as:

(+1)+(+5)+3(x) = 0

x = -2

The oxidation number of O in $O_{2}$ is Zero (o).

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Answer:

See figure 1

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See figure 1

I hope it helps!

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