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garik1379 [7]
2 years ago
12

How many particles of oxygen gas are in 22.0 grams of oxygen ?

Chemistry
1 answer:
lianna [129]2 years ago
5 0

Moles of oxygen = 22g / 32 (Mr of O2)

Moles of oxygen = 0.7 moles (1 dp)

Avogadro's constant = 6.02×10^23 particles per mole

0.7 moles×(6.02×10^23) = Answer

This is assuming it wants O2, if it wants pure oxygen then replace 32 with 16.

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Explain, in terms of electrons, why the radius of a calcium ion is smaller than the radius of a calcium atom.
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3 years ago
What is the vapor pressure of CS2CS2 in mmHgmmHg at 26.5 ∘C∘C? Carbon disulfide, CS2CS2, has PvapPvap = 100 mmHgmmHg at −−5.1 ∘C
Digiron [165]

Answer: 26.5 mm Hg

Explanation:

The vapor pressure is determined by Clausius Clapeyron equation:

ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1= initial pressure at 26.5^oC = ?

P_2 = final pressure at -5.1^oC = 100 mm Hg

= enthalpy of vaporisation = 28.0 kJ/mol =28000 J/mol

R = gas constant = 8.314 J/mole.K

T_1= initial temperature = 26.5^oC=273+26.5=299.5K

T_2 = final temperature =-5.1^oC=273+(-5.1)=267.9K

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

\log (\frac{P_1}{100})=\frac{28000}{2.303\times 8.314J/mole.K}[\frac{1}{299.5}-\frac{1}{267.9}]

\log  (\frac{P_1}{100})=-0.576

\frac{P_1}{100}=0.265

P_1=26.5mmHg

Thus the vapor pressure of CS_2CS_2 in mmHg at 26.5 ∘C is 26.5

7 0
3 years ago
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