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bekas [8.4K]
4 years ago
9

At 25°c the henry's law constant for nitrogen trifluoride (nf3) gas in water is 7.9 × 10-4 m/atm. what is the mass of nf3 gas th

at can be dissolved in 150 ml of water at 25°c and an nf3 partial pressure of 1.71 atm?
Chemistry
1 answer:
Step2247 [10]4 years ago
4 0
For this problem, we should use the Henry's Law formula which is written below:

P = kC
where
P is the partial pressure of the gas
k is the Henry's Law constant at a certain temperature
C is the concentration

Substituting the values,
1.71 atm = (7.9×10⁻⁴<span> /atm)C
Solving for C,
C = 2164.56 molal or 2164.56 mol/kgwater

Let's make use of density of water (</span>1 kg/1 m³) and the molar mass of NF₃ (71 g/mol).<span>

Mass of NF</span>₃ = 2164.56 mol/kg water * 1 kg/1 m³ * 1 m³/1000000 mL * 150 mL * 71 g/mol = 23.05 g 
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Answer:

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Explanation:

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4 years ago
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The density of lava is 3.1 g/cm². It has a volume of 2700 cm3. How much does this block of lava weigh?
konstantin123 [22]

Answer:

Mass of lava is 8370 g.

Explanation:

Density:

Density is equal to the mass of substance divided by its volume.

Units:

SI unit of density is Kg/m3.

Other units are given below,

g/cm3, g/mL , kg/L

Formula:

D=m/v

D= density

m=mass

V=volume

Symbol:

The symbol used for density is called rho. It is represented by ρ. However letter D can also be used to represent the density.

Given data:

density of lava = 3.1 g/cm³

volume= 2700 cm³

mass= ?

Solution:

d = m/v

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m = 8370 g

4 0
3 years ago
How many moles are present in 10.5 x 10^23 atoms of helium?
Usimov [2.4K]

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3 years ago
A bond between two iodine atoms would have what as the percent ionic character
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7. How much energy in KJ is released when 15.0 g of steam at 100.0 degrees Celsius is condensed and then frozen to ice at 0.0 de
Aliun [14]

Answer:

-6.27 kj

Explanation:

Given data:

Energy released = ?

Mass of steam = 15 g

Initial temperature = 100°C

Final temperature = 0°C

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Specific heat capacity of water is 4.18 j/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT =  0°C - 100°C = -100°C

Q = 15g × 4.18 j/g.°C  × -100°C

Q = -6270 j

J to KJ:

-6270 j/1000 = -6.27 kj

4 0
4 years ago
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