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Delvig [45]
2 years ago
12

A weather balloon is filled with helium that occupies a volume of 5.37 104 L at 0.995 atm and 32.0°C. After it is released, it r

ises to a location where the pressure is 0.720 atm and the temperature is -11.7°C. What is the volume of the balloon at that new location?
Chemistry
1 answer:
svp [43]2 years ago
8 0

Answer:

The new volume is 63583 L

Explanation:

Step 1: Data given

The initial volume of the balloon = 5.37 * 10^4 L

The initial pressure = 0.995 atm

The initial temperature = 32.0 °C = 305.15 K

The pressure decreased to 0.720 atm

The temperature decreased to -11.7 °C = 261.45 K

Step 2: Calculate the new volume

P1*V1 / T1 = P2*V2/T2

⇒with P1 = the initial pressure = 0.995 atm

⇒with V1 = the initial volume = 5.37 *10^4 L

⇒with T1 = the initial temperature = 305.15 K

⇒with P2 = the decreased pressure = 0.720 atm

⇒with V2 = the new volume = TO BE DETERMINED

⇒with T2 = the decreased temperature : 261.45 K

(0.995 * 5.37*10^4)/305.15 = (0.720 * V2) / 261.45

V2 = 63583 L

The new volume is 63583 L

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They are good conductors of heat and electricity.

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Answer: Balanced equation that is formed by combining these two half reactions is Cu + NO^{-}_{3} + 2H^{+} \rightarrow Cu^{2+} + NO^{-}_{2} + H_{2}O.

Explanation:

A chemical equation which contains same number of atoms on both reactant and product side is called a balanced chemical equation.

For example, Cu \rightarrow Cu^{2+} + 2e^{-}

NO^{-}_{3} + 2e^{-} + 2H^{+} \rightarrow NO^{-}_{2} + H_{2}O

On cancelling the common species from both these half-reactions, the complete balanced equation will be as follows.

Cu + NO^{-}_{3} + 2H^{+} \rightarrow Cu^{2+} + NO^{-}_{2} + H_{2}O

Thus, we can conclude that balanced equation that is formed by combining these two half reactions is Cu + NO^{-}_{3} + 2H^{+} \rightarrow Cu^{2+} + NO^{-}_{2} + H_{2}O.

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3 years ago
One of the reactions that occurs in a blast furnace, in which iron ore is converted to cast iron, is Fe2O3 + 3CO → 2Fe + 3CO2 Su
Tpy6a [65]

Answer : The percent purity of Fe_2O_3 in the original sample is 87.94 %

Explanation :

The given balanced chemical reaction is:

Fe_2O3+3CO\rightarrow 2Fe+3CO_2

First we have to calculate the mass of Fe.

\text{Moles of }Fe=\frac{\text{Mass of }Fe}{\text{Molar mass of }Fe}

Molar mass of Fe = 55.8 g/mole

\text{Moles of }Fe=\frac{1.79\times 10^3kg}{55.8g/mole}=\frac{1.79\times 10^3\times 1000g}{55.8g/mole}=3.15\times 10^4mole

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From the balanced chemical reaction we conclude that,

As, 2 moles of Fe produced from 1 mole of Fe_2O_3

So, 3.15\times 10^4mole of Fe produced from \frac{3.15\times 10^4}{2}=15750 mole of Fe_2O_3

Now we have to calculate the mass of Fe_2O_3

\text{ Mass of }Fe_2O_3=\text{ Moles of }Fe_2O_3\times \text{ Molar mass of }Fe_2O_3

Molar mass of Fe_2O_3 = 159.69 g/mole

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Now we have to calculate the percent purity of Fe_2O_3 in the original sample.

Mass of original sample = 2.86\times 10^3kg

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\text{Percent purity}=\frac{2.515\times 10^3kg}{2.86\times 10^3kg}\times 100=87.94\%

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