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DerKrebs [107]
3 years ago
4

If you mix different amounts of two ideal gases that are originally at different temperatures, what must be true of the final st

ate after the temperature stabilizes? (There may be more than one correct choice.)
A. Both gases will reach the same final temperature.
b) The final rms molecular speed will be the same for both gases.c) The final average kinetic energy of a molecule will be the same for both gases.
Chemistry
1 answer:
lara [203]3 years ago
6 0

Answer:

a) Both gases will reach the same final temperature.

c) The final average kinetic energy of a molecule will be the same for both gases.

Explanation:

The molecules of the hotter gas will have higher average kinetic energies than those of the cooler gas. As the molecules of the molecules of the two gases collide with each other and exchange energies, the gases will reach an equilibrium in which their molecules have the same average kinetic energy.

Since average kinetic energy is a measure of temperature, both gases will reach the same final temperature.

b) is wrong. The rms speeds will differ, depending on the molecular masses of the two gases.

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4. How many grams of ammonium carbonate are needed to decompose in order to produce
Thepotemich [5.8K]

Answer:

14.23g of (NH4)2CO3

Explanation:

We'll begin by writing the balanced equation for the reaction.

(NH4)2CO3 –> (NH4)2O + CO2

Next,, we shall determine the mass of (NH4)2CO3 that decomposed and the mass of CO2 produced from the balanced equation. This is illustrated below:

Molar mass of (NH4)2CO3 = 2[14+(4x1)] + 12 + (16x3)

= 2[14 +4] + 12 + 48

= 2[18] + 60 = 96g/mol

Mass of (NH4)2CO3 from the balanced equation = 1 x 96 = 96g

Molar mass of CO2 = 12 + (2x16) = 44g/mol

Mass of CO2 from the balanced equation = 1 x 44 = 44g.

Summary:

From the balanced equation above,

96g of (NH4)2CO3 decomposed to produce 44g of CO2.

Finally, we can determine the mass of (NH4)2CO3 that decomposed to produce 6.52g of CO2 as follow:

From the balanced equation above,

96g of (NH4)2CO3 decomposed to produce 44g of CO2.

Therefore, Xg of (NH4)2CO3 will decompose to produce 6.52g of CO2 i.e

Xg of (NH4)2CO3 = (96 x 6.52)/44

Xg of (NH4)2CO3 = 14.23g

Therefore, 14.23g of (NH4)2CO3 is needed to produce 6.52g of CO2.

4 0
3 years ago
How much water would be needed to completely dissolve 1.52 L of the gas at a pressure of 730 torr and a temperature of 21 ∘C?
Sophie [7]

Answer:

The correct answer is 0.4 L.

Explanation:

The mentioned question can be solved by using the equation,  

C = K × Pgas--------(i)

Here K is the Henry law constant whose value is 0.158 mol/L/atm, C is the concentration of the gas in liquid state, and Pgas is the partial pressure of the gas.  

Now to find the volume of water, the formula to be used is,  

PV = nRT-----------(ii)

Here P is the pressure of the gas, V is the volume, R is the universal gas constant whose value is 0.082 Latm/mol/K and T is the temperature.  

PgasV = nRT

Pgas = nRT/Vgas

The value of Pgas is inserted in equation (i) we get,  

C = K × nRT/Vgas

It is to be noted that C = n/V, here n is the no. of the moles and V is the volume of liquid.  

n/Vliquid = K × nRT/Vgas

1/Vliquid = KRT/Vgas

Vliquid = Vgas/KRT--------------(iii)

Based on the given information, the volume of the gas is 1.52 L, the value of K is 0.158 mol/L/atm, the value of R is 0.082 Latm/mol/K and value of T is 21 degree C or 273 + 21 = 294 K.  

Now putting the values in equation (iii) we get,  

Vliquid = 1.52 L / 0.158 × 0.082 × 294

Vliquid = 1.52 / 3.809

Vliquid = 0.399 or 0.4 L

Hence, the volume of water required to dissolve 1.52 L of gas is 0.4 L.  

8 0
3 years ago
The value of Henry's law constant for oxygen in water at 242C is 1.66 x 1 ox M/torr. b. c. Calculate the solubility of oxygen in
DochEvi [55]
25ca an the oxygen gas is 2 atom ur mum
4 0
3 years ago
Help please hello please
lord [1]

Answer:

the

first one

Explanation:

5 0
3 years ago
Read 2 more answers
What are 2 applications of the Colligative Property that allows freezing point depression?
butalik [34]
1) Adding salt on the roads in the winter time,
2) Adding salt to the ice bath when making ice cream.
3) Mixing 2 substances when making coolant for  a car.
5 0
3 years ago
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