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timofeeve [1]
2 years ago
7

Uranium hexafluoride is a solid at room temperature, but it boils at 56C. Determine the density of uranium hexafluoride at 60.C

and 745 torr.
Chemistry
1 answer:
Mandarinka [93]2 years ago
5 0

Answer: The density of Uranium Hexafluoride at 60 °C and 745 torr is 0.0127 g/ml

Explanation:

Uranium Hexafluoride is only present as gas at 60 °C and 745 torr.

So, we know PV = nRT, from that

n/V = P/ RT = (745 torr)/(62.36 L-torr/mol*K)(337 K) = 0.0354 mol/L.

Density = (352.02 g/mol)(0.0354 mol/L)/(1000 cm³/L) = 0.0127 g/cm³

To learn more about density of Uranium Hexafluoride from the given link

brainly.com/question/17987457

#SPJ4

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Explain mole concept using any real life example
andrew-mc [135]

Answer:

for instance :

water , H²O , and hydrogen peroxide . H²O² , are alike

in that their respective molecules are composed of hydrogen and oxygen atomd . hope its helpful . Good luck :)

5 0
2 years ago
A very hot cube of copper metal (32.5 g) is submerged into 105.3 g of water at 15.4 0C and it reach a thermal equilibrium of 17.
zysi [14]

Answer:

The initial temperature of the metal is 84.149 °C.

Explanation:

The heat lost by the metal will be equivalent to the heat gain by the water.  

- (msΔT)metal = (msΔT)water

-32.5 grams × 0.365 J/g°C × ΔT = 105.3 grams × 4.18 J/g °C × (17.3 -15.4)°C

-ΔT = 836.29/12.51 °C

-ΔT = 66.89 °C

-(T final - T initial) = 66.89 °C

T initial = 66.89 °C + T final

T initial = 66.89 °C + 17.3 °C

T initial = 84.149 °C.

7 0
2 years ago
Help me answer this question please?
Tema [17]

Answer:

I think D

Explanation:

Ok, I'm not sure but it sounds right ish you should check a practice video or something. It might also be B or C but im pretty certain it isnt A just ask yourself is the student measuring it in newtons? Is that important in the process? What about if the student is considering the affect of mass is it important? Good luck srry if im not much of help! If this is like A SUPER IMPORTANT TEST OR SOMETHING RLLLLLLLY IMPORTANT just wait for another answer gl!

5 0
2 years ago
0.500 L of a gas is collected at 2911 MM and 0°C. What will the volume be at STP?
ioda

Answer:

V₂ =  1.92 L

Explanation:

Given data:

Initial volume = 0.500 L

Initial pressure =2911 mmHg (2911/760 = 3.83 atm)

Initial temperature = 0 °C (0 +273 = 273 K)

Final temperature = 273 K

Final volume = ?

Final pressure = 1 atm

Solution:

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

by putting values,

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 3.83 atm × 0.500 L × 273 K / 273 K × 1 atm

V₂ = 522.795 atm .L. K / 273 K.atm

V₂ =  1.92 L

4 0
3 years ago
Dinitrogen tetraoxide, N2O4, decomposes to nitrogen dioxide, NO2, in a first-order process. If k = 1.5 x 103 s-1 at 5 ºC and k =
Arada [10]

Answer:

The activation energy for the decomposition = 33813.28 J/mol

Explanation:

Using the expression,

\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )

Wherem  

k_1\ is\ the\ rate\ constant\ at\ T_1

k_2\ is\ the\ rate\ constant\ at\ T_2

E_a is the activation energy

R is Gas constant having value = 8.314 J / K mol  

Thus, given that, E_a = ?

k_2=4.0\times 10^3s^{-1}

k_1=1.5\times 10^3s^{-1}  

T_1=5\ ^0C  

T_2=25\ ^0C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (5 + 273.15) K = 278.15 K  

T = (25 + 273.15) K = 298.15 K  

T_1=278.15\ K

T_2=298.15\ K

So,

\ln \frac{1.5\times 10^3}{4.0\times 10^3}\:=-\frac{E_{a}}{8.314}\times \left(\frac{1}{278.15}-\frac{1}{298.15}\right)

E_a=-\ln \frac{1.5\times \:10^3}{4.0\times \:10^3}\:\times \frac{8.314}{\left(\frac{1}{278.15}-\frac{1}{298.15}\right)}

E_a=-\frac{8.314\ln \left(\frac{1.5\times \:10^3}{4\times \:10^3}\right)}{\frac{1}{278.15}-\frac{1}{298.15}}

E_a=-\frac{689483.53266 \ln \left(\frac{1.5}{4}\right)}{20}

E_a=33813.28\ J/mol

<u>The activation energy for the decomposition = 33813.28 J/mol</u>

8 0
3 years ago
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