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3 years ago

7

Suppose that 0.00150 moles of CO2 (44.0 g/mol) effuse out of a pinhole in 1.00 hour. How many moles of N2 (28.0 g/mol) would eff

use out of the same pinhole in 1.00 hour? Assume both gases are at the same temperature and pressure.

2 answers:

kenny6666 [7]3 years ago

4 0

Answer:

I was having trouble with this question idk the answer

Explanation:

ololo11 [35]3 years ago

3 0

Answer:

0.0019mol

Explanation:

To solve this problem, first we calculate for the rates of the effusion:

For CO2,

Number of mole = 0.00150 mol

Time = 1h

R1 = mol / time

R1 = 0.0015/1

R1 = 0.0015mol/h

Molar Mass of CO2 (M1) = 44.0 g/mol

For N2,

R2 =?

Molar Mass of N2 (M2) = 28.0 g/mol

Using Graham's law, we can obtain Rate (R2) of N2 as follows:

R1/R2 = √(M2/M1)

0.0015/R2 = √(28/44)

0.0015/R2 = 0.7977

R2 = 0.0015 /0.7977

R2 = 0.0019mol/h

But rate = mole /time

Time = 1h

Rate od N2 = 0.0019mol/h

0.0019 = mole /1

Mole = 0.0019mol

Therefore, the number of mole of N2 is 0.0019mol

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BartSMP [9]

Answer:

Odds to be given for an event that either Romance or Downhill wins is 11:4

Explanation:

Given an odd, r = a : b. The probability of the odd, r can be determined by;

Pr(r) = $\frac{a}{b}$ ÷ (

So that;

Odd that Romance will win = 2:3

Pr(R) = $\frac{2}{3}$ ÷ (

= $\frac{2}{3}$ ÷ $\frac{5}{3}$

= $\frac{2}{5}$

Odd that Downhill will win = 1:2

Pr(D) = $\frac{1}{2}$ ÷ (

= $\frac{1}{2}$ ÷ $\frac{3}{2}$

= $\frac{1}{3}$

The probability that either Romance or Downhill will win is;

Pr(R) + Pr(D) = $\frac{2}{5}$ + $\frac{1}{3}$

= $\frac{11}{15}$

The probability that neither Romance nor Downhill will win is;

Pr(neither R nor D) = (1 - $\frac{11}{15}$ )

= $\frac{4}{15}$

The odds to be given for an event that either Romance or Downhill wins can be determined by;

= Pr(Pr(R) + Pr(D)) ÷ Pr(neither R nor D)

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= $\frac{11}{4}$

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8 0

3 years ago

When a 2.00 g sample of KCl is dissolved in water in a calorimeter that has a total heat capacity of 1.28 kJ ⋅ K − 1 , the tempe

Alchen [17]

Answer : The molar heat of solution of KCl is, 17.19 kJ/mol

Explanation :

First we have to calculate the heat of solution.

$q=c\times (\Delta T)$

where,

q = heat produced = ?

c = specific heat capacity of water = $1.28kJ/K$

$\Delta T$ = change in temperature = 0.360 K

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

$q=1.28kJ/K\times 0.360K$

$q=0.4608kJ=460.8J$

Now we have to calculate the molar heat solution of KCl.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat released = 460.8 J

m = mass of $KCl$ = 2.00 g

Molar mass of $KCl$ = 74.55 g/mol

$\text{Moles of }KCl=\frac{\text{Mass of }KCl}{\text{Molar mass of }KCl}=\frac{2.00g}{74.55g/mole}=0.0268mole$

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

$\Delta H=\frac{460.8J}{0.0268mole}$

$\Delta H=17194.029J/mol=17.19kJ/mol$

Therefore, the molar heat of solution of KCl is, 17.19 kJ/mol

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andreyandreev [35.5K]

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I hope this helps! :)

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3 years ago