Answer:
Part 1) 85.3 grams NaCl
Part 2) 8.79 x 10²³ formula units NaCl
Explanation:
<u>(Part 1)</u>
To find the mass of NaCl, you need to multiply the given value (1.46 moles) by the molar mass of NaCl. This measurement is the atomic masses of the elements times each of their quantities combined. In this case, there is only one mole of each element in the molecule. Moles should be located in the denominator of the conversion to allow for the cancellation of units. The final answer should have 3 sig figs to reflect the given value.
Molar Mass (NaCl): 22.99 g/mol + 35.45 g/mol
Molar Mass (NaCl): 58.44 g/mol
1.46 moles NaCl 58.44 g
--------------------------- x ---------------- = 85.3 grams NaCl
1 mole
<u>(Part 2)</u>
I do not know which other question the second part is referring to, so I will just use the moles given in the first part. To find the formula units, you need to multiply the given value (1.46 moles NaCl) by Avogadro's Number. This conversion represents the number of formula units found in 1 mole of the sample. The moles should be in the denominator of the conversion to allow for the cancellation of units.
Avogadro's Number:
1 mole = 6.022 x 10²³ formula units
1.46 moles NaCl 6.022 x 10²³ units
------------------------ x ----------------------------- = 8.79 x 10²³ formula units NaCl
1 mole
Answer : The vapor pressure of bromine at 
is 0.1448 atm.
Explanation :
The Clausius- Clapeyron equation is :
where,
= vapor pressure of bromine at = ?
= vapor pressure of propane at normal boiling point = 1 atm
= temperature of propane = 
= normal boiling point of bromine = 
= heat of vaporization = 30.91 kJ/mole = 30910 J/mole
R = universal constant = 8.314 J/K.mole
Now put all the given values in the above formula, we get:
Hence, the vapor pressure of bromine at is 0.1448 atm.
