Molar mass Na = 23.0 g/mol
1 mol ---- 23.0 g
n mol ---- 69 g
n = 69 / 23.0
n = 3.0 moles
1 mole -------- 6.02x10²³ molecules
3.0 moles ---- ?
3.0 * 6.02x10²³ / 1
= 1.806x10²⁴ molecules
hope this helps!
The powder sugar because has more contact area
Answer is: molality od sodium chloride is 2,55 mol/kg.
V(solution) = 100 ml.
m(solution) = d(solution) · V(solution).
m(solution) = 1,10 g/ml · 100 ml.
m(solution) = 110 g.
ω(NaCl) = 13,0% = 0,13.
m(NaCl) = ω(NaCl) · m(solution).
m(NaCl) = 0,13 · 110 g.
m(NaCl) = 14,3 g.
n(NaCl) = m(NaCl) ÷ M(NaCl).
n(NaCl) = 14,3 g ÷ 58,5 g/mol.
n(NaCl) = 0,244 mol.
m(H₂O) = 110 g - 14,3 g.
m(H₂O) = 95,7 g = 0,0957 kg.
b(NaCl) = n(NaCl) ÷ m(H₂O).
b(NaCl) = 0,244 mol ÷ 0,0957 kg.
b(NaCl) = 2,55 mol/kg.
The experimental mole ratio of silver chloride to barium chloride is calculated as below
fin the mole of each compound
mole= mass/molar mass
moles of AgCl = 14.5g/142.5 g/mol = 0.102 moles of AgCl
moles of BaCl2 = 10.2 g/208 g/mol = 0.049 moles of BaCl2
find the mole ratio by dividing each mole with the smallest mole(0.049)
AgCl= 0.102/0.049 =2
BaCl2 = 0.049/0.049 =1
therefore the mole ratio AgCl to BaCl2 is 2 :1
Answer: The atomic weight of the metal would be 85.47.
Explanation:
Mass of isotope 1 of metal = 84.9118
% abundance of isotope 1 of metal = 72.15% =
Mass of isotope 2 of metal= 86.9092
% abundance of isotope 2 of metal = 27.85% =
Formula used for average atomic mass of an element :

Therefore, the atomic weight of the metal would be 85.47.