Atomic mass number is the number of protons and neutrons. Subtract 80-35=45 is the number of protons. Because the atom is neutrally charged, the number of protons must equal the number of electrons, so there are 45 electrons.
0.091 moles are contained in 2.0 L of N2 at standard temperature and pressure.
Explanation:
Data given:
volume of the nitrogen gas = 2 litres
Standard temperature = 273 K
Standard pressure = 1 atm
number of moles =?
R (gas constant) = 0.08201 L atm/mole K
Assuming nitrogen to be an ideal gas at STP, we will use Ideal Gas law
PV = nRT
rearranging the equation to calculate number of moles:
PV = nRT
n = 
putting the values in the equation:
n = 
n = 0.091 moles
0.091 moles of nitrogen gas is contained in a container at STP.
Answer:
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The equation for the reaction between NaOH and AlCl₃ is as follows;
3NaOH + AlCl₃ ---> 3NaCl + Al(OH)₃
the stoichiometry of NaOH : AlCl₃ is 3:1
3 moles of NaOH reacts with 1 mol of AlCl₃ to produce 1 mol of Al(OH)₃
the number of AlCl₃ moles reacted - 6.5 mol
molar mass of NaOH -(23 +16 +1) = 40 g/mol
the number of NaOH moles reacted = 57.0 g / 40 g/mol
NaOH moles = 1.425 mol
either NaOH or AlCl₃ is in excess and other is the limiting reactant.
limiting reactant is the reactant whose number of moles are fully consumed during the reaction. the reactant that is in excess will have leftover moles that are remaining after the reaction.
If AlCl₃ is the limiting reactant, number of NaOH moles would be thrice the amount of AlCl₃ present,
then number of NaOH moles that should be present - 6.5 * 3 = 19.5 mol
however there are only 1.425 mol of NaOH present, therefore AlCl₃ is in excess.
Then NaOH is the limiting reactant,
the amount of products formed depends on the amount of the limiting reactant present.
stoichiometry of NaOH : Al(OH)₃ is 3:1
the number of Al(OH)₃ moles produced = number of NaOH moles reacted / 3
number of Al(OH)₃ moles are - 1.425 mol /3 = 0.475 mol
molar mass of Al(OH)₃ = (27 +3*16 + 3*1) = 78 g/mol
mass of Al(OH)₃ produced = 78 g/mol * 0.475 mol = 37.05 g
