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charle [14.2K]
3 years ago
11

If two gases with pressures of 2 atm and 3 atm are mixed at a constant temperature, what will the total pressure be?

Chemistry
2 answers:
slamgirl [31]3 years ago
8 0
If two gases with pressures of 2 atm and 3 atm are mixed at constant temperature, the total pressure will be the sum of the two pressures. Therefore the answer is D. 2 atm + 3 atm or 5 atm will be the total pressure of the gas mixture. 
kotegsom [21]3 years ago
6 0

Answer:

D. 2 atm + 3 atm

Explanation:

Given:

Pressure Gas 1 = 2 atm

Pressure Gas 2 = 3 atm

Explanation:

As per Dalton's Law in a mixture of gases, the total pressure is the sum of the partial pressures of the individual gases.

Ptotal = P1 + P2 + P3 + ...

where P1, P2,P3 are the partial pressures

Therefore for the given gas mixture, the total pressure would be:

Ptotal = 2 atm + 3 atm = 5 atm

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describe the photoelectric effect and explain why it made modifications to the Rutherford model necessary
spayn [35]

<u>Answer:</u>

photoelectric effect and the modifications to the Rutherford model  is necessary

<u>Explanation:</u>

Photoelectric effect refers to emission of electrons when a light of certain frequency falls into a metal surface or a metal atom.

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In this reaction, a gaseous system with a volume of 3.00 L has a rate of formation of NOCl of 0.0120 M s-1 . The volume of the s
oee [108]

Answer:

x = 0.324 M s⁻¹

Explanation:

Equation for the reaction can be represented as:

 2 NO(g)   + Cl₂ (g)      ⇄     2NOCl (g)

Rate = K [NO]² [Cl₂]

Concentration = \frac{numbers of mole (n)}{volume (v)}

from the question; their number of moles are constant since the species are quite alike.

As such; if Concentration varies inversely proportional to the volume;

we have: Concentration ∝ \frac{1}{v}

Concentration = \frac{1}{v}

Similarly; the Rate can now be expressed as:

Rate = K [NO]² [Cl₂]

Rate = (\frac{1}{v}) ^2 (\frac{1}{v} )

Rate = \frac{1}{v^3}

We were also told that the in the reaction, the gaseous system has an initial volume of 3.00 L and rate of formation of 0.0120 Ms⁻¹

So we can have:

0.0120 = \frac{1}{3^3}

0.0120 = \frac{1}{27}   -----Equation (1)

Now; the new rate of formation when the  volume of the system decreased to 1.00 L can now be calculated as:

x = (\frac{1}{1})^3

x = 1             ------- Equation (2)

Dividing equation (2) with equation (1); we have:

\frac{0.0210}{x} = \frac{\frac{1}{27} }{1}

\frac{0.0210}{x} = \frac{1}{27}

x = 0.0120 × 27

x = 0.324 M s⁻¹

∴  the new rate of formation of NOCl = 0.324 M s⁻¹

8 0
3 years ago
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