The concentration is expressed in molarity or moles solute per liters solution. The molar mass of ethanol is 46 g/mol. To obtain a 1.2 L of vodka, the following stoichiometric calculation is done:
6.86 mol/L * 1.2 L * (46g/mol) = 378.672 g of ethanol is needed
Flammability
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First calculate number of Nitrogen atoms contained by 1 mole of Zn(NO₃)₂,
As one formula unit of Zn(NO₃)₂ contains 2 Nitrogen atoms, then one mole of Zn(NO₃)₂ will contain,
= 2 × 6.022 × 10²³
= 1.20 × 10²⁴ Atoms of Nitrogen / mole
Now, calculate number of Nitrogen Atoms in 0.150 moles of Zn(NO₃)₂.
As,
1 mole of Zn(NO₃)₂ contained = 1.20 × 10²⁴ atoms of N
Then,
0.15 mole of Zn(NO₃)₂ will contain = X atoms of N
Solving for X,
X = (0.15 mol × 1.20 × 10²⁴ atom) ÷ 1 mol
X = 1.80 × 10²⁴ Atoms of Nitrogen
Result:
So, 0.15 mole of Zn(NO₃)₂ contains 1.80 × 10²⁴ atoms of Nitrogen.
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Answer: Molecule
Step-by-step-explanation
