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MArishka [77]
3 years ago
10

Determine the mass in grams of 125 mol of neon

Chemistry
1 answer:
Vesnalui [34]3 years ago
5 0
Molar mass of Neon ( Ne ) = 20.1797 g/mol

m = n * mm

m = 125 * 20.1797

m = 2522.4625 g

hope this helps!

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RideAnS [48]

Answer:

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8 0
3 years ago
The reaction is run at two temperatures where temperature 1 is lower than temperature 2. Which relationship is correct for eithe
Tasya [4]

Answer:

The answer is "K_1 < K_2"

Explanation:

In the given question, the value of the K=Ae^{-\frac{Ea}{RT}} and the T_1is the rate value which is the constant that is  K_2>K_1. As per the temperature value when its increase rate is constantly increasing. E_a is activation energy it is not dependent on the temperature that why the answer is K_1 < K_2.

8 0
3 years ago
What is the temperature of a gas that is expanded from 3.75 L at 37 degrees Celsius to 5.6 L?
deff fn [24]

Answer:

190 °C  

Step-by-step explanation:

The pressure is constant, so this looks like a case where we can use <em>Charles’ Law</em>:  

V₁/T₁ = V₂/T₂      Invert both sides of the equation.  

T₁/V₁ = T₂/V₂      Multiply each side by V₂

T₂ = T₁ × V₂/V₁

=====

V₁ = 3.75 L; T₁ = (37 + 273.15) K = 310.15 K  

V₂ = 5.6 L;   T₂ = ?  

=====

T₂ = 310.15 × 5.6/3.75

T₂ = 310.15 × 1.49

T₂ = 463 K

t₂ = 463 – 273.15

t₂ = 190 °C

3 0
3 years ago
Calcium oxide or quicklime (CaO) is used in steelmaking, cement manufacture, and pollution control. It is prepared by the therma
Elena-2011 [213]

Answer:

The yearly release of CO_2 into the atmosphere is 6.73\times 10^{10} kg.

Explanation:

CaCO_3(s)\rightarrow CaO(s) + CO_2(g)

Annual production of CaO = 8.6\times 10^{10} kg=8.6\times 10^{13} g

Moles of CaO :

\frac{8.6\times 10^{13} g}{56 g/mol}=1.53\times 10^{12} moles

According to reaction, 1 mole of CaO is produced along with 1 mole of carbon-dioxide.

Then along with  1.53\times 10^{12} moles of CaO moles of carbon-dioxide moles produced will be:

\frac{1}{1}\times 1.53\times 10^{12} moles=1.53\times 10^{12} moles of carbon-dioxide

Mass of 1.53\times 10^{12} moles of carbon-dioxide:

1.53\times 10^{12}mol\times 44 g/mol=6.73\times 10^{13} g =6.73\times 10^{10} kg

The yearly release of CO_2 into the atmosphere is 6.73\times 10^{10} kg.

6 0
3 years ago
Based on the following molecular weight data of polypropylene, determine the degree of polymerization Molecular Weight Range (g/
Lady bird [3.3K]

Answer:

785

Explanation:

Molecular. X. W

Weight

8000-16000 0.05 0.03

16000-24000. 0.017. 0.08

24000-32000. 0.22. 0.18

32000-40000. 0.25. 0.35

40000-48000. 0.22. 0.27

48000-56000. 0.09. 0.09

Mean weight X*M. W*M

12000. 600. 240

20000. 3200. 2000

28000. 6720. 5600

36000. 10080. 10800

44000. 8800. 11880

52000. 3640 3640

Total=33040g\mol 36240

Note before repeat molecular weight m= 3*12.01+6*1.008=

42.08g/mol

Degree of polymerization= total W*M/w=33040/42.08 =785

8 0
3 years ago
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