A 5.15L balloon has a pressure of 1.35 atm. If compressed to 3.43L,

Chemistry

1 answer:

Semenov [28]2 years ago

4 0

Solution:

Using Ideal Gas law:

P1 = 1.45 atm

V1 = 5.15 L

P2 = ?

V2 = 3.43 L

Using Ideal Gas Law, PV = nRT:

Now, the number of moles (n) is the same independent of pressure and distance, so let's say that the temperature is stable since there is no details on the issue. As a consequence,

P1V1 = nRT = P2V2

P1V1 = P2V2

Solve for P2 and you're going to get 2.03 atm. It makes sense as you decrease the amount of the gas by compressing it, which implies that you raise the heat.

2) P = 1.5 atm

T = 301 K

n = 0.785 moles

PV = nRT

V = nRT/P

R = 0.0821 (L*atm)/(mol * K)

Therefore,

V = (0.785 * 0.0821 * 301)/1.5 = 19.4 L

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For the reaction KClO2⟶KCl+O2 KClO2⟶KCl+O2 assign oxidation numbers to each element on each side of the equation. K in KClO2:K i

frosja888 [35]

Answer:

Explanation:

The formula of the reaction:

KClO₂ → KCl + O₂

To assign oxidation numbers, we have to obey some rules:

Elements in an uncombined state or one whose atoms combine with one another to form molecules have an oxidation number of zero.
The charge on simple ions signifies their oxidation number.
The algebraic sum of all the oxidation number of all atoms in a neutral compound is zero. For radicals with charges, their oxidation number is the charge.

The oxidation number of K in KClO₂:

K + (-1) + 2(-2) = 0

K-5 = 0

K = +5

The oxidation number of K in KCl:

K + (-1) = 0

K = +1

The oxidation number Cl in KClO₂ is -1

For Cl in KCl, the oxidation number is -1

For O in KClO₂, the oxidation number is (2 x -2) = -4

For O in O₂, the oxidation number is 0

K moves from an oxidation state of +5 to +1. This is a gain of electrons and K has undergone reduction. We then say K is reduced.