2BF₃ + 3Li₂SO₃ ----> B₂(SO₃)₃ + <u>6LiF
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1) Formulas:
a) mole fraction of component 1, X1
X1 = number of moles of compoent 1 / total number of moles
b) Molar mass = number grams / number of moles => number of moles = number of grams / molar mass
2) Application
Number of moles of CaI2 = 0.400
Molar mass of water = 18.0 g/mol
Number of moles of water: 850.0 g / 18.0 g/mol = 47.22 mol
Total number of moles = 0.400 + 47.22 =47.62
Molar fraction of CaI2 = 0.400 / 47.62 = 0.00840
Answer:
D. its actual size and distance from an observer
Explanation:
haha aint i the brainliest
jst kidding
hope it helped u
Answer is: 2) 117g.
2Na + Cl₂ → 2NaCl
Step 1: calculate amount of substance of sodium and chlorine.
n(Na) = m(Na)÷M(Na) = 46g ÷ 23 g/mol = 2 mol.
n(Cl₂) = m(Cl₂)÷M(Cl₂) = 71g ÷ 71 g/mol = 1 mol.
Step 2: calculate amount of substance and mass of sodium-chloride.
Because both sodium and chlorine react completely, we can use both n to compare with n of NaCl.
n(Na) : n(NaCl) = 2:2, 2 mol : n(NaCl) = 2:2
n(NaCl) = 2mol, m(NaCl) = 2mol ·5805 g/mol = 117 g.
