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Iteru [2.4K]
3 years ago
7

Three 5-l flasks, fixed with pressure gauges and small valves, each contains 4 g of gas at 273 k. flask a contains h2, flask b c

ontains he, and flask c contains ch4. rank the flask contents in terms of

Chemistry
1 answer:
Varvara68 [4.7K]3 years ago
7 0
First, please check the missing part in your question in the attachment.
a) So first, the Rank of pressure:
according to this formula PV = nRT and when n = m/Mw
PV = m/Mw * R*T
when we have the same mass m and the same V volume so P will proportional with the mole weight M as when the M is smaller the pressure will be greater 
when Mw of H2(A) = 2 g / Mw of He (B) = 4 g and Mw of CH4(C) = 16 g
∴ Pressure :
 (A) > (B) > (c)

B) The rank of average molecular kinetic energy:
when K = 3/2 KB T
when K is the average kinetic energy per molecule of gas 
and KB is Boltzmann's constant
and T is the temperature (K)
So from this equation, we can know that K only depends on T value, and when we have the T constant here for A, B, and C So the rank of K will be like the following:
∴ A = B = C
C) the rank of diffusion rate after the valve is opened:
according to this formula:
R2/R1 = √M1/M2
from this equation, we can see that diffusion is proportional to the reciprocal of the molecular mass M so,
when Mw H2 (A) = 2 g & Mw He(B) = 4 g & CH4 (C) = 16 g
∴ the rank of diffusion:
A > B > C

D) The rank of the Total kinetic energy of the molecules:
when we have the Mw different so it will make the no.of molecules differs as when the Mw is low the no.of molecules will be hight, and when the average molecular kinetic energy equals. so the total kinetic energy will depend on no. of molecules 
∵ Mw A < Mw B < Mw C 
∴no .of molecules of A > B >C
∴ the rank of total kinetic energy is:
A > B > C

e) the rank of density:

when ρ = m/ v 
and m is the mass & v is the volume and we have both is the same for A, B, and C
so the density also will be the same, ∴ the rank of the density is:
A = B = C

F) the rank of the collision frequency:
as the no.of molecules increase the collision frequency increase and depend also on the velocity and it's here the same.
∴ Collision frequency will only depend on the no.of molecules
we have no.of molecules of A > B > C as Mw A < B < C 
∴the rank of the collision frequency is:
A > B > C 

 



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zavuch27 [327]

Answer:

13.00 g

Explanation:

To solve this problem, you need to understand the law of conservation of mass.  This law means that whatever amount you start out with is what you should have at the end.  The amount may be in different forms, but that amount is still there.

If you produced 10.00 g of B and 3.00 g of A, you should have started out with 13.00 g of AB.

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8 0
4 years ago
1) Why does the spinning motion cool the wet bulb thermometer in a sling psychrometer when there is less humidity in the air?
disa [49]

Answer:1)Once the wick is wet, hold the handle and whirl the psychrometer in the air. Whirling generates airflow around both the wet and dry bulbs. This action allows water in the saturated wick to evaporate, thereby reducing the temperature of the wet-bulb thermometer.

2)The rate of evaporation from the wet bandage on the bulb, and the temperature difference between the dry bulb and wet bulb, depends on the humidity of the air. The evaporation is reduced when the air contains more water vapor.

Explanation:

6 0
2 years ago
A lab team places 0.250g of an unknown solid acid in an Erlenmeyer flask. They neutralize the solid acid with 15.0ml of 0.210 M
Otrada [13]

Answer:

<u>Molar</u><u> </u><u>mass</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>unknown</u><u> </u><u>acid</u><u> </u><u>is</u><u> </u><u>7</u><u>9</u><u> </u><u>grams</u>

Explanation:

We have to first get moles in 15.0 ml of sodium hydroxide solution:

{ \sf{1 \: l \: of \:NaOH \: contains \: 0.210 \: moles }} \\ { \sf{0.015 \: l \: of \: NaOH \: contain \: (0.015 \times 0.210) \: moles }} \\  { \underline{ = 0.00315 \: moles \: of \: NaOH}}

since mole ratio of acid : base is 1 : 1, so;

moles of acid that reacted is <u>0</u><u>.</u><u>0</u><u>0</u><u>3</u><u>1</u><u>5</u><u> </u><u>m</u><u>o</u><u>l</u><u>e</u><u>s</u><u> </u><u>o</u><u>f</u><u> </u><u>t</u><u>h</u><u>e</u><u> </u><u>u</u><u>n</u><u>k</u><u>n</u><u>o</u><u>w</u><u>n</u><u> </u><u>a</u><u>c</u><u>i</u><u>d</u><u>.</u>

then we've to get molar mass:

{ \sf{0.00315 \: moles \: of \: acid \: weigh \: 0.250 \: g}} \\ { \sf{1 \: mole \: of \: acid \: weighs \: ( \frac{1 \times 0.250}{0.00315} ) \: g}} \\ { \underline{ = 79.4 \: g \approx79 \: grams}}

6 0
3 years ago
a solution with a transmittance of 0.44 is analyzed in a spectrophotometer with 6% stray light. calculate the absorbance reporte
IgorLugansk [536]

The absorbance reported by the defective instrument was 0.3933.

Absorbance A = - log₁₀ T

Tm = transmittance measured by spectrophotometer

Tm = 0.44

Absorbance reported in this equipment = -log₁₀ (0.44) = 0.35654

True absorbance can be calculated by true transmittance, Tm = T+S(α-T)

S = fraction of stray light = 6%= 6/100 = 0.06

α= 1, ideal case

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Tm = T+S(α-T)

now, T= Tm-S/ 1-S = 0.44-0.06/ 1-0.06 = 0.404233

therefore, actual reading measured is A = -log₁₀ T = -log₁₀ (0.404233)

i.e; 0.3933

To know more about transmittance click here:

brainly.com/question/17088180

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2 years ago
Can someone please help me out with my homework. I don't understand this at all and it's due tomorrow 1st period. Thx
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5.

shown in image above.

6 0
3 years ago
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