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kirill115 [55]
3 years ago
14

Solve the problem. Express your answer to the correct number of significant figures

Chemistry
1 answer:
Shalnov [3]3 years ago
4 0

Answer:

Explanation:

30.53131

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You have a gas at a volume of 9.71 L at a pressure of 209 torr and at 10.1 °C. What
Usimov [2.4K]

Answer:

219.95 °C

Explanation:

Given data:

Volume of gas = 9.71 L

Initial pressure = 209 torr (209/760 = 0.275 atm)

Initial temperature = 10.1 °C (10.1 +273 = 283.1 K)

Final temperature = ?

Final pressure = 364 torr (364/760 =0.479 atm)

Solution:

According to Gay-Lussac Law,

The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.

Mathematical relationship:

P₁/T₁ = P₂/T₂

Now we will put the values in formula:

0.275 atm / 283.1 K = 0.479 atm/T₂

T₂ = 0.479 atm × 283.1 K/ 0.275 atm

T₂ = 135.6 atm. K /0.275 atm

T₂ = 493.1 K

Kelvin to °C:

493.1 K - 273.15 = 219.95 °C

8 0
2 years ago
The science of molecules and and their is called chemistry
GarryVolchara [31]
Is this supposed to be a question?????
6 0
3 years ago
Read 2 more answers
What is the mole fraction of solute in a 3.19 m aqueous solution?
Bumek [7]
Molality= mol/ Kg

if we assume that we have 1 kg of water, we have 3.19 moles of solute. 

the formula for mole fraction --> mole fraction= mol of solule/ mol of solution

1) if we have 1 kg of water which is same as 1000 grams of water. 

2) we need to convert grams to moles using the molar mass of water 

molar mass of H₂O= (2 x 1.01) + 16.0 = 18.02 g/mol

1000 g (1 mol/ 18.02 grams)= 55.5 mol

3) mole of solution= 55.5 moles + 3.19 moles= 58.7 moles of solution

4) mole fraction= 3.19 / 58.7= 0.0543 
4 0
3 years ago
The reaction C4H8(g)⟶2C2H4(g) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ/mol.262 kJ/mol. At 600.0 K,600.0 K, the rate c
crimeas [40]

Answer : The rate constant at 785.0 K is, 1.45\times 10^{-2}s^{-1}

Explanation :

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

or,

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = rate constant at 600.0K = 6.1\times 10^{-8}s^{-1}

K_2 = rate constant at 785.0K = ?

Ea = activation energy for the reaction = 262 kJ/mole = 262000 J/mole

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 600.0K

T_2 = final temperature = 785.0K

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

\log (\frac{K_2}{6.1\times 10^{-8}s^{-1}})=\frac{262000J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{785.0K}]

K_2=1.45\times 10^{-2}s^{-1}

Therefore, the rate constant at 785.0 K is, 1.45\times 10^{-2}s^{-1}

7 0
3 years ago
Which of the following is an example of a parasite?
snow_tiger [21]
C
a parasite is an organism that lives inside a host and benefits from taking away nutrients at the others expense
7 0
2 years ago
Read 2 more answers
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