Answer:
1. 9.57 × 10^-9 moles.
2. 7.38mol
Explanation:
1.) To find the number of moles there are in the number of particles in an atom, we divide the number of particles (nA) by Avagadro's constant (6.02 × 10^23)
Hence, to find the number of moles (n) of Manganese (Mn), we say:
5.76 x 10^15 atoms ÷ 6.02 × 10^23
5.76/6.02 × 10^(15-23)
= 0.957 × 10^-8
= 9.57 × 10^-9 moles.
2.) Mole = mass/molar mass
Molar mass of sodium chloride (NaCl) = 23 + 35.5
= 58.5g/mol
mole = 431.6 g ÷ 58.5g/mol
mole = 7.38mol
<u>Answer:</u> The experimental van't Hoff factor is 1.21
<u>Explanation:</u>
The expression for the depression in freezing point is given as:
where,
i = van't Hoff factor = ?
= depression in freezing point = 0.225°C
= Cryoscopic constant = 1.86°C/m
m = molality of the solution = 0.100 m
Putting values in above equation, we get:
Hence, the experimental van't Hoff factor is 1.21
Answer:
they both have 4 sig figs:)
Explanation:
The pairs are:
K, Kr - Same period
Be, Mg - Same group
Ni, Tc - Both are transition metals
B, Ge - Both are metaloids
Al, Pb - Both form inert oxides
