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Ymorist [56]
3 years ago
7

What effect will the number of particles and temperature have on the rate of diffusion of gas molecules?

Chemistry
1 answer:
xxTIMURxx [149]3 years ago
7 0

Answer: option B - The rate of diffusion increases with the increase in number of particles and temperature.

Explanation:

Gases belong to the states of matter. They are made up of particles loosely attached, and traveling through the entire space of their containing vessel. These travels is known as DIFFUSION.

An INCREASE in TEMPERATURE and PARTICLES would DEFINITELY increase rate of TRAVEL (as particles tends to escape through any available space).

For Example: You must have noticed that on blowing more air into a BALLOON, It tends to burst. So, also heating a gas containing vessel, it EXPANDS and soon burst.

So, the RIGHT ANSWER is option B

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Thousand of molecules are called​
kozerog [31]

Answer:

monomers

Explanation:

Thousands of molecules are called monomers

6 0
3 years ago
Calculate the root means square velocity of nitrogen molecules in 25 degrees Celsius
Mademuasel [1]
Calculate the root mean square velocity of nitrogen molecules at 25°C.
297 m/s
149 m/s
515 m/s
729 m/s
7 0
3 years ago
Help??
kvasek [131]

Answer:

Mass of sodium chloride decomposed = 24.54 g

Explanation:

Given data:

Mass of sodium chloride decomposed = ?

Mass of chlorine gas formed = 15 g

Solution:

Chemical equation:

2NaCl      →         2Na + Cl₂

Number of moles of Cl₂:

Number of moles = mass/molar mass

Number of moles = 15 g/ 71 g/mol

Number of moles = 0.21 mol

Now we will compare the moles of Cl₂ with NaCl from balance chemical equation.

                    Cl₂            :              NaCl

                      1              :                2

                      0.21         :            2×0.21 = 0.42 mol

Mass of Sodium chloride decompose:

Mass = number of moles × molar mass

Mass = 0.42 mol × 58.44 g/mol

Mass = 24.54 g

4 0
3 years ago
Carbon tetrachloride, CCl4, was once used as a dry cleaning solvent, but is no longer used because it is carcinogenic. At 57.8 °
Mila [183]
This problem is to use the Claussius-Clapeyron Equation, which is:

ln [p2 / p1] = ΔH/R [1/T2 - 1/T1]

Where p2 and p1 and vapor pressure at estates 2 and 1

ΔH is the enthalpy of vaporization

R is the universal constant of gases = 8.314 J / mol*K

T2 and T1 are the temperatures at the estates 2 and 1.

The  normal boiling point => 1 atm (the pressure of the atmosphere at sea level) = 101,325 kPa

Then p2 = 101.325 kPa
T2 = ?
p1 = 54.0 kPa
T1 = 57.8 °C + 273.15K = 330.95 K
ΔH = 33.05 kJ/mol = 33,050 J/mol 

=> ln [101.325/54.0] = [ (33,050 J/mol) / (8.314 J/mol*K) ] * [1/x - 1/330.95]

=> 0.629349 = 3975.22 [1/x - 1/330.95] = > 1/x =  0.000157 + 1/330.95 = 0.003179

=> x = 314.6 K => 314.6 - 273.15 = 41.5°C

Answer: 41.5 °C 
3 0
3 years ago
the temperature of a sample of water increases from 20celsius to 46.6celsius as it absorbs 5650 J of heat. what is the mass of t
Levart [38]

Answer:

m = 50.74 kg

Explanation:

We have,

Initial temperature of water is 20 degrees Celsius

Final temperature of water is 46.6 degrees Celsius

Heat absorbed is 5650 J

It is required to find the mass of the sample. The heat absorbed is given by the formula ad follows :

Q=mc\Delta T

c is specific heat of water, c = 4.186 J/g°C

So,

m=\dfrac{Q}{c\Delta T}\\\\m=\dfrac{5650}{4.186\times (46.6-20)}\\\\m=50.74\ kg

So, the mass of the sample is 50.74 kg.

8 0
3 years ago
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