Answer:
molality = 0.564 m
Explanation:
Molality = number of moles of solute / kg of solvent
1- getting moles of solute:
number of moles = mass / molar mass
we have:
mass = 373.5 g
molar mass = 331.2 g/mol
Therefore:
number of moles = 373.5 / 331.2 = 1.128 moles
2- getting kg of solvent:
mass in kg = mass in grams * 10⁻³
mass in kg = 2 * 10³ * 10⁻³
mass in kg = 2 kg
3- getting molality:
Molality = number of moles of solute / kg of solvent
Molality = 1.128 / 2
Molalty = 0.564 m
Hope this helps :)
Answer:
1,033.56 grams of carbon dioxide was emitted into the atmosphere.
Explanation:
Energy absorbed by pork,E =
(assuming)
Total energy produced by barbecue = Q
Percentage of energy absorbed by pork = 10%


Since, it is a energy produced in order to indicate the direction of heat produced we will use negative sign.
Q = 
Moles of propane burnt to produce Q energy =n


According to reaction , 1 mol of propane gives 3 moles of carbon dioxide. then 7.83 moles of will give:
carbon dioxide gas.
Mass of 23.49 moles of carbon dioxide gas:
23.49 mol × 44 g/mol =1,033.56 g
1,033.56 grams of carbon dioxide was emitted into the atmosphere.
Na because its a metal. Metals are the best conductors. S and Ne are nonmetals. and Ge is a metalloid. (Metalloids are semi conductors)