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olchik [2.2K]
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.
Chemistry
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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QUESTION THREE
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)

                               = \frac{11}{15} ÷ \frac{4}{15}

                              = \frac{11}{4}

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

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

7 0
3 years ago
Which of the following measurements contains 2 significant figures?
Trava [24]
The answer would be 371 because it has multiple complete digits
7 0
3 years ago
Why the mass of a rusted nail is greater then the nail before it rusted?
kifflom [539]
This is the reaction formula,
4Fe+3O2=2Fe2O3
<span>3Fe+202=Fe3O4
it has o</span>xygen atom after it's rusted
6 0
3 years ago
What are the stable isotopes of carbon?
andreyandreev [35.5K]
Answer:

The stable isotopes of carbon are carbon-12 (12C) and carbon-13 (13C).

I hope this helps! :)
3 0
3 years ago
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