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alexandr1967 [171]
2 years ago
6

An ideal gas has a pressure of 2.5 atm, a volume of 1.0 L at a temperature of 30°C. How many molecules are there in this gas? (R

= 8.31 J/mol ∙ K,1.00 atm = 101 kPa, NA = 6.022 × 1023)
Chemistry
1 answer:
Molodets [167]2 years ago
3 0

Answer:

In the gas there are 6.02x10²² molecules

Explanation:

This is a problem to be solved by the Ideal Gas Law formula, which is this one:

Pressure . volume = n° moles . R . T° in K

The excersise presented R as 8.31 J/mol, but If i follow the others data, I'd rather use 0.082 L.atm / mol.K

Let's replace numbers:

2.5 atm . 1L = n° moles. 0.082L.atm / mol.K . 303K

(2.5 atm . 1L) / (0.082 mol.K/L.atm . 303K) = n° moles

0.100 moles = n

Now that we have the moles, we can use NA to calculate the total of molecules:

1 mol ___ has __ NA molecules

0.1 mol has ____ ( 0.1 . 6.02x10²³) = 6.02x10²² molecules

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Suppose the half-life is 9.0 s for a first order reaction and the reactant concentration is 0.0741 M 50.7 s after the reaction s
bazaltina [42]

<u>Answer:</u> The time taken by the reaction is 84.5 seconds

<u>Explanation:</u>

The equation used to calculate half life for first order kinetics:

k=\frac{0.693}{t_{1/2}}

where,

t_{1/2} = half-life of the reaction = 9.0 s

k = rate constant = ?

Putting values in above equation, we get:

k=\frac{0.693}{9}=0.077s^{-1}

Rate law expression for first order kinetics is given by the equation:

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}     ......(1)

where,

k = rate constant  = 0.077s^{-1}

t = time taken for decay process = 50.7 sec

[A_o] = initial amount of the reactant = ?

[A] = amount left after decay process =  0.0741 M

Putting values in equation 1, we get:

0.077=\frac{2.303}{50.7}\log\frac{[A_o]}{0.0741}

[A_o]=3.67M

Now, calculating the time taken by using equation 1:

[A]=0.0055M

k=0.077s^{-1}

[A_o]=3.67M

Putting values in equation 1, we get:

0.077=\frac{2.303}{t}\log\frac{3.67}{0.0055}\\\\t=84.5s

Hence, the time taken by the reaction is 84.5 seconds

6 0
3 years ago
CK-12 Boyle and Charles's Laws if Mrs. Pa pe prepares 12.8 L of laughing gas at 100.0 k Pa and -108 °C and then she force s the
nirvana33 [79]

Answer:

The answer to your question is   P2 = 2676.6 kPa

Explanation:

Data

Volume 1 = V1 = 12.8 L                        Volume 2 = V2 = 855 ml

Temperature 1 = T1 = -108°C               Temperature 2 = 22°C

Pressure 1 = P1 = 100 kPa                    Pressure 2 = P2 =  ?

Process

- To solve this problem use the Combined gas law.

                     P1V1/T1 = P2V2/T2

-Solve for P2

                     P2 = P1V1T2 / T1V2

- Convert temperature to °K

T1 = -108 + 273 = 165°K

T2 = 22 + 273 = 295°K

- Convert volume 2 to liters

                       1000 ml -------------------- 1 l

                         855 ml --------------------  x

                         x = (855 x 1) / 1000

                         x = 0.855 l

-Substitution

                    P2 = (12.8 x 100 x 295) / (165 x 0.855)

-Simplification

                    P2 = 377600 / 141.075

-Result

                   P2 = 2676.6 kPa

3 0
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