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pishuonlain [190]
3 years ago
8

An air tight freezer measures 4 mx 5 m x 2.5 m high. With the door open, it fills with 22 °C air at 1 atm pressure.

Chemistry
1 answer:
____ [38]3 years ago
7 0

Answer:

(a) Density of the air = 1.204 kg/m3

(b) Pressure = 93772 Pa or 0.703 mmHg

(c) Force needed to open the door =  15106 N

Explanation:

(a) The density of the air at 22 deg C and 1 atm can be calculated using the Ideal Gas Law:

\rho_{air}=\frac{P}{R*T}

First, we change the units of P to Pa and T to deg K:

P=1 atm * \frac{101,325Pa}{1atm}=101,325 Pa\\\\ T=22+273.15=293.15^{\circ}K

Then we have

\rho_{air}=\frac{P}{R*T}=\frac{101325Pa}{287.05 J/(kg*K)*293.15K} =1.204 \frac{kg}{m3}

(b) To calculate the change in pressure, we use again the Ideal Gas law, expressed in another way:

PV=nRT\\P/T=nR/V=constant\Rightarrow P_{1}/T_{1}=P_{2}/T_{2}\\\\P_{2}=P_{1}*\frac{T_{2}}{T_{1}}=101325Pa*\frac{7+273.15}{22+273.15}=101,325Pa*0.9254=93,772Pa\\\\P2=93,772 Pa*\frac{1mmHg}{133,322Pa}= 0.703 mmHg

(c) To calculate the force needed to open we have to multiply the difference of pressure between the inside of the freezer and the outside and the surface of the door. We also take into account that Pa = N/m2.

F=S_{door}*\Delta P=2m^{2} *(101325Pa - 93772Pa)=2m^{2} *7553N/m2=15106N

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the reversible reaction N2(g) 3H2(g) &lt;--&gt; 2NH3(g) produces ammonia, which is a fertilizer. At equilibrium, a 1L flask cont
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Explanation:

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3 years ago
Identify which sets of quantum numbers are valid for an electron. Note that each set is ordered (n,l,ml,ms).
Fynjy0 [20]

Answer:

The answer to your question is:

Explanation:

Quantum numbers have the numbers

n = from 1 to 7

l = s = 0 p = 1 d = 2  f= 3

m = ± l

s = ±1/2

Group of answer choice.

A) 2, -1. 1, -1/2     The option is incorrect "l" could not be negative

B) 2, 1, 0 , 1/2       The option is posible

C) 3, 2, 1, -1          This option is incorrect "s" can not value -1

D) 4, 3, 2, 1/2       This option is correct

E) 4, 3, -4, 1/2      This option is incorrect "m2 can not value -4

F) 1, 1, 0, 1/2         This option is incorrect, the second number is not posible

G) 1, 1, 0, -1/2        This option is incorrect the second number could not be 1.

H) 3, 0, 0, 1/2       This option is incorrect "l" can not value 0

I) 0, 2, 0, 1/2         This option is incorrect, n can not value 0

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3 0
3 years ago
calculate the water potential of a solution of 0.15m sucrose. the solution is at standard temperature.
Mrac [35]

Answer:

The water potential of a solution of 0.15 M sucrose solution is -3.406 bar.

Explanation:

Water potential = Pressure potential + solute potential

P_w=P_p+P_s

P_w=P_p+(-iCRT)

We have :

C = 0.15 M, T = 273.15 K

i = 1

The water potential of a solution of 0.15 m sucrose= P_w

P_p=0 bar (At standard temperature)

P_s=-iCRT=-\times 1\times 8.314\times 10^{-2}bar L/mol K\times 273.15 K=-3.406 bar

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The water potential of a solution of 0.15 M sucrose solution is -3.406 bar.

7 0
3 years ago
47.9 ml hrdrogen is collected at 26° Celsius and 718 torr. Find the volume occupied at STP
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Answer:

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Explanation:

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Where P = Initial pressure of hydrogen, V = Initial volume of hydrogen, T= Initial Temperature of hydrogen, P' = Final pressure of hydrogen, V' = Final Volume of Hydrogen, T' = Final Temperature.

make V' the subject of the equation

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V' = ( 95725.196×0.0479×273)/(299×101000)

V' = 0.04145 dm³

V' = 41.45 mL

4 0
3 years ago
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