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frez [133]
3 years ago
13

Assuming that the container is completely full, that the temperature is 22.1 ∘C, and that the atmospheric pressure is 1.1 atm ,

calculate the percent (by volume) of air that would be displaced if all of the liquid nitrogen evaporated. (Liquid nitrogen has a density of 0.807 g/mL.) Express your answer using two significant figures.
Chemistry
1 answer:
Kisachek [45]3 years ago
5 0

Answer : The percent (by volume) of air displaced if all of the liquid nitrogen evaporated would be 29 %

Explanation :

Let us assume that 1.2 L container of liquid nitrogen is kept in a closet measuring 1.0m by 1.3m by 2.0m.

First we have to calculate the mass of nitrogen.

Density=\frac{Mass}{Volume}

Given:

Density of nitrogen = 0.807 g/mL

Volume of nitrogen = 1.2 L = 1200 mL    (1 L = 1000 mL)

0.807g/mL=\frac{Mass}{1200mL}

Mass=0.807g/mL\times 1200mL=968.4g

Now we have to calculate the moles of nitrogen.

\text{Moles of }N_2=\frac{\text{Mass of }N_2}{\text{Molar mass of }N_2}

Molar mass of nitrogen (N_2) = 28 g/mole

\text{Moles of }N_2=\frac{968.4g}{28g/mole}=34.58mole

Now we have to calculate the volume of nitrogen.

As we know that,

PV = nRT

where,

P = pressure of nitrogen = 1.1 atm

V = volume of nitrogen = ?

n = number of moles of nitrogen = 34.58 mole

R = gas constant = 0.0821 L.atm/mol.K

T = temperature of nitrogen = 22.1^oC=273+22.1=295.1K

Now put all the given values in this formula, we get:

1.1atm\times V=34.58mole\times 0.0821L.atm/mol.K\times 295.1K

V=761.6L

Now we have to calculate the volume of container.

The formula of volume of cuboid, we use the equation:

V=lbh

where,

V = volume of container = ?

l = length of container = 1.0 m

b = breadth of container = 1.3 m

h = height of container = 2.0 m

Putting values in above equation, we get:

V=1.0m\times 1.3m\times 2.0m=2.6m^3=2600L

Conversion used : 1m^3=1000L

Now we have to calculate the percent (by volume) of air that would be displaced if all of the liquid nitrogen evaporated.

\% \text{ of displaced volume}=\frac{761.6L}{2600L}\times 100=29.29\% \approx 29\%

Therefore, the percent (by volume) of air displaced if all of the liquid nitrogen evaporated would be 29 %

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Calculate the density of oxygen,
abruzzese [7]

Answer:

density of oxygen = 1.307 g/l

Explanation:

given pressure = 1 atm temperature = 298 k

PM=DRT

where p is pressure M is molecular mass D is density R is gas constant and T is temperature

substituting the values in the equation

1\times 32=D\times 0.0821\times 298

D = 1.307 g/l.

5 0
3 years ago
Any reaction that absorbs 150 kcal of energy can be classified as
Dennis_Churaev [7]
The answer should be (D) endothermic reaction.

Reason : -  Reactions in which energy is absorbed are called endothermic reactions. In your provided question, 150 kcal energy is being <u>absorbed</u>, and thus, we can say that it is an endothermic reaction. 
6 0
3 years ago
Calculate the equilibrium constant for the decomposition of water 2h2o(l)  2h2(g) + o2(g) at 25°c, given that g°f (h2o(l)) = –
kow [346]

Answer:

2.6 ×10^-42

Explanation:

From

∆G= -RTlnK

∆G= -237.2 KJmol-1 or -237.2×10^3 Jmol-1

R= 8.314 Jmol-1K-1

T= 25°C + 273= 298K

-237.2×10^3= 8.314 × 298 × ln K

ln K= -237.2×10^3/2477.572

K = 2.6 ×10^-42

3 0
3 years ago
What is the empirical formula of a compound that is 24.42 % calcium, 17.07 % nitrogen, and 58.5% oxygen?
REY [17]

Answer:

CaN_{2} O_{6}

Explanation:

When calculating an empirical formula from percentages, assume you have a 100g sample. This allows you to convert the percentages directly to grams, because X % of 100g is X grams.

So:

24.42 % = 24.42 g Ca, 17.07% = 17.07g N, 58.5% = 58.5g O

The next step is to divide each mass by their molar mass to convert your grams to moles.

24.42/40.08 = 0.6092 mol

17.07/14.01 = 1.218 mol

58.85/15.99 = 3.680 mol

Then you will divide all of your mol values by the SMALLEST number of moles. This gives you whole numbers that are the mole ratio (subcripts) of the empircal formula.

0.6092 mol/0.6092 mol = 1

1.218 mol/0.6092 mol = 2

3.680 mol/0.6092 mol = 6

So the empirical formula is CaN_{2} O_{6}

5 0
2 years ago
If 1.0 gram of hydrogen reacts with 19.0 grams of fluorine, then what is the percent by mass of fluorine in the compound that is
astraxan [27]
The reaction between hydrogen (H2) and fluorine (F2) is given below,
                                   H2 + F2 ---> 2HF
One mole of both hydrogen and fluorine yields to 2 moles of hydrogen fluoride. This can also be expressed as, 2 grams of hydrogen and 38 grams of fluorine will form 40 grams of hydrogen fluoride. From the given, only 20 grams of HF is formed with 19 g of it being fluorine. Thus, the percentage fluorine of the compound formed is 95%. 
8 0
3 years ago
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