What will happen to the balloon and cloth if they are brought close together without touching? Explain.
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The answer for the following question is mentioned below.
<u><em>Therefore no of moles present in the gas are 1.12 moles</em></u>
Explanation:
Given:
Pressure of gas (P) = 1.2 atm
Volume of a gas (V) = 50.0 liters
Temperature (T) =650 K
To calculate:
no of moles present in the gas (n)
We know;
According to the ideal gas equation;
We know;
<u>P × V = n × R × T </u>
where,
P represents pressure of the gas
V represents volume of the gas
n represents no of the moles of a gas
R represents the universal gas constant
where the value of R is 0.0821 L atm mole^{-1} K^-1
T represents the temperature of the gas
As we have to calculate the no of moles of the gas;
n =
n = \frac{1.2*50.0}{0.0821*650}
n = \frac{60}{53.365}
n = 1.12 moles
<u><em>Therefore no of moles present in the gas are 1.12 moles</em></u>
S₂O₈²⁻
(aq) + 2I⁻
(aq) → I₂(aq) + 2SO₄
²⁻(aq)
2S₂O₃²⁻
(aq) + I₂(aq) → S₄O₆²⁻
(aq) + 2I⁻
(aq)
<u>Explanation:</u>
S₂O₈²⁻
(aq) + 2I⁻
(aq) → I₂(aq) + 2SO₄
²⁻(aq)
To measure the rate of this reaction we must measure the rate of concentration change of one of the reactants or products. To do this, we will include (to the reacting S₂O₈
²⁻ and I⁻
i) a small amount of sodium thiosulfate, Na₂S₂O₃,
ii) some starch indicator.
The added Na₂S₂O₃ does not interfere with the rate of above reaction, but it does consume the I₂ as soon as it is formed.
2S₂O₃²⁻
(aq) + I₂(aq) → S₄O₆²⁻
(aq) + 2I⁻
(aq)
This reaction is much faster than the previous, so the conversion of I2 back to I⁻ is essentially instantaneous.