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omeli [17]
2 years ago
7

Nitrogen gas was collected by water displacement. What was pressure of the N, collected if the temperature was 22 °C?​

Chemistry
1 answer:
Anestetic [448]2 years ago
5 0

The pressure of the N₂ collected if the temperature was 22°C celcius is 98.6823 kPa.

What is atmospheric pressure?

The air around you has weight, and it presses against everything it touches. That pressure is called atmospheric pressure, or air pressure.

Atmospheric pressure = 101.325 kPa

Vapor pressure of water at 22°C = 2.6427 kPa

Pressure of N₂ = (Atmospheric pressure) - (vapour pressure of water at 22°C)

Pressure of N₂ = (101.325 kPa) - ( 2.6427 kPa)

Pressure of N₂ = 98.6823 kPa

Therefore, the pressure of the N₂ collected if the temperature was 22°C celcius is 98.6823 kPa.

Learn more about atmospheric pressure here:

brainly.com/question/6492699

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Describe how you can figure out how many grams of a COMPOUND if you know how many moles there
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Divide the mass of the compound in grams by the molar mass you just calculated. The answer is the number of moles of that mass of compound. For example, 25 grams of water equals 25/18.016 or 1.39 moles.
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What is the balanced equation for the reaction of aqueous cesium sulfate and aqueous barium perchlorate?
Aleksandr-060686 [28]

Answer:

The balanced chemical reaction is given as:

Cs_2SO_4(aq)+Ba(ClO_4)_2(aq)\rightarrow BaSO_4(s)+2CsClO_4(aq)

Explanation:

When aqueous cesium sulfate and aqueous barium perchlorate are mixed together it gives white precipitate barium sulfate and aqueous solution od cesium perchlorate.

The balanced chemical reaction is given as:

Cs_2SO_4(aq)+Ba(ClO_4)_2(aq)\rightarrow BaSO_4(s)+2CsClO_4(aq)

According to reaction, 1 mole of cesium sulfate reacts with 1 mole of barium perchlorate to give 1 mole of a white precipitate of barium sulfate and 2 moles of cesium perchlorate.

3 0
3 years ago
The balanced redox reactions for the sequential reduction of vanadium are given below.
Minchanka [31]

Answer : 0.0392 grams of Zn metal would be required to completely reduced the vanadium.

Explanation :

Let us rewrite the given equations again.

2VO_{2}^{+} (aq)+ 4H^{+}(aq) + Zn (s)\rightarrow 2VO^{2+}(aq)+Zn^{2+}+2H2O(l)2VO^{2+}(aq)+ 4H^{+}(aq) + Zn (s)\rightarrow 2V^{3+} (aq)+Zn^{2+}+2H2O(l)

2V^{3+} (aq)+ Zn (s)\rightarrow 2V^{2+}(aq)+Zn^{2+}(aq)

On adding above equations, we get the following combined equation.

2VO_{2}^{+} (aq)+ 8H^{+} (aq) + 3Zn (s)\rightarrow 2V^{2+}(aq)+3Zn^{2+}(aq)+4H_{2}O(l)

We have 12.1 mL of 0.033 M solution of VO₂⁺.

Let us find the moles of VO₂⁺ from this information.

12.1 mL \times \frac{1L}{1000mL}\times \frac{0.033mol}{L}=0.0003993mol NO_{2}^{+}

From the combined equation, we can see that the mole ratio of VO₂⁺ to Zn is 2:3.

Let us use this as a conversion factor to find the moles of Zn.

0.0003993mol NO_{2}^{+}\times \frac{3mol Zn}{2molNO_{2}^{+}}=0.00059895mol Zn

Let us convert the moles of Zn to grams of Zn using molar mass of Zn.

Molar mass of Zn is 65.38 g/mol.

0.00059895mol Zn\times \frac{65.38gZn}{1molZn}=0.0392gZn

We need 0.0392 grams of Zn metal to completely reduce vanadium.

6 0
3 years ago
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