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Mekhanik [1.2K]
3 years ago
14

If 1.6 moles of an ideal gas are at a temperature of 25c and a pressure of 0.91 atm hg, what volume would the gas occupy, in lit

ers?
Chemistry
1 answer:
Fiesta28 [93]3 years ago
8 0
From\ Ideal\ Gas\ Law:\\\\PV=nRT\\\\
P-pressure\\V-volume\\n-number\ of\ moles\\R- universal\ gas\ constant\\T-absolute\ tmeperature\\\\
R=\ 8,3145J/ mol\ K\\P=0,91atm\\
n=1,6moles\\T=25c\\\\
Converting\ units:\\\\
1atm=101,3kPa=101,3*10^3\\
0,91atm=0,91*101,3*10^3=92,183*10^3Pa\\T=25c=273+25=298K\\\\
V=\frac{nRT}{P}=\frac{1,6*8,3145*298}{92183}=0,043m^3=43liters\\\\Answer\ is\ 43\ liters.
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How many molecules of CaCl2 are equivalent to 75.9 g CaCl2
sergij07 [2.7K]
First, you need to find:
One mole of CaCl_{2} is equivalent to how many grams?

Well, for this you have to look up the periodic table. According to the periodic table:
The atomic mass of Calcium Ca = 40.078 g (See in group 2)
The atomic mass of <span>Chlorine Cl = 35.45 g (See in group 17)
</span>
As there are two atoms of Chlorine present in CaCl_{2}, therefore, the atomic mass of CaCl_{2} would be:

Atomic mass of CaCl_{2}  = Atomic mass of Ca + 2 * Atomic mass of Cl

Atomic mass of CaCl_{2} = 40.078 + 2 * 35.45 = 110.978 g

Now,

110.978 g of CaCl_{2} = 1 mole.
75.9 g of CaCl_{2} = \frac{75.9}{110.978} moles = 0.6839 moles.

Hence,
The total number of moles in 75.9g of CaCl_{2} = 0.6839 moles

According to <span>Avogadro's number,
1 mole = 1 * </span>6.022 * 10^{23} molecules
0.6839 moles = 0.6839 * 6.022 * 10^{23} molecules = 4.118*10^{23} molecules

Ans: Number of molecules in 75.9g of  CaCl_{2} =  4.118*10^{23} molecules

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