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VashaNatasha [74]
2 years ago
9

(Hurry ASAP!!!) There are 1.5 moles of methane (CH4) in a tank. It has a pressure of 4.4 atm at 173°C. Find the volume occupied

by the gas.
Chemistry
1 answer:
slamgirl [31]2 years ago
8 0

Answer:

hope this help !

Explanation:

Use the given functions to set up and simplify 173 ° C .

1.5 =

CH4 = CH4

4.4 = CH4

173 ° C = CH4

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Andrew [12]
The two angles would be both 41 degrees.
(2x  + 13) = 41 \\ 2x = 41 - 13 \\ 2x = 28 \\ x =  \frac{28}{2} \\ x = 14
x in this case would be 14 degrees.
4 0
3 years ago
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The temperature inside your freezer is 0 degrees Celsius. You place a balloon with an initial temperature of 30 degrees C and a
blsea [12.9K]

Answer:

V=0.68L

Explanation:

For this question we can use

V1/T1 = V2/T2

where

V1 (initial volume )= 0.75 L

T1 (initial temperature in Kelvin)= 303.15

V2( final volume)= ?

T2 (final temperature in Kelvin)= 273.15

Now we must rearrange the equation to make V2 the subject

V2= (V1/T1) ×T2

V2=(0.75/303.15) ×273.15

V2=0.67577931717

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4 0
3 years ago
How many kilocalories are needed to vaporize 5.8 mol of Br2
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<span>The vaporization of br2 from liquid to gas state requires 7.4 k/cal /mol.</span>
8 0
2 years ago
Two moles of an ideal gas are placed in a container whose volume is 2.3 x 10^-3 m3. The absolute pressure of the gas is 6.9 x 10
PtichkaEL [24]

Answer:

K.E.=1.97\times 10^{-21}\ J

Explanation:

Given that:-

Pressure = 6.9\times 10^5\ Pa

The expression for the conversion of pressure in Pascal to pressure in atm is shown below:

P (Pa) = \frac {1}{101325} P (atm)

Given the value of pressure = 43,836 Pa

So,  

6.9\times 10^5\ Pa = \frac{6.9\times 10^5}{101325} atm

Pressure = 6.80977 atm

Volume = 2.3\times 10^{-3}\ m^3 = 2.3 L ( 1 m³ = 1000 L)

n = 2 mol

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

6.80977 atm × 2.3 L = 2 mol × 0.0821 L.atm/K.mol × T

⇒T = 95.39 K

The expression for the kinetic energy is:-

K.E.=\frac{3}{2}\times K\times T

k is Boltzmann's constant = 1.38\times 10^{-23}\ J/K

T is the temperature

So, K.E.=\frac{3}{2}\times 1.38\times 10^{-23}\times 95.39\ J

K.E.=1.97\times 10^{-21}\ J

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2 years ago
“Enter the conjugate acids of no2- and no3-“
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Answer:

nitrit,nitrat

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