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.
Answer:
No, tobacco companies do not reuse tobacco in their cigarettes, I also did some extra research to make sure I wasn’t giving a false answer
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A- Identify the mixture:
The mixture of powdered charcoal and powdered sugar is considered as a homogeneous mixture. This means that you cannot identify the components with naked eye as they are uniformly distributed in the mixture.
B- Separate components:
You ca separate the charcoal powder from the sugar powder using the following steps:
1- add water. Sugar will dissolve in water while charcoal won't.
2- filter the solution where the powdered charcoal will remain on the filter paper and the solution of powder will pass through.
3- boil the sugar solution (above 100 degrees celcius). The water will evaporate and the sugar will precipitate.
Answer:
molecules C6H12O6 = 2.674 E22 molecules.
Explanation:
from periodic table:
⇒ molecular mass C6H12O6 = ((6)(12.011)) + ((12)(1.008)) + ((6)(15.999))
⇒ Mw C6H12O6 = 180.156 g/mol
⇒ mol C6H12O6 = (8.00 g)(mol/180.156 g) = 0.0444 mol C6H12O6
∴ mol ≡ 6.022 E23 molecules
⇒ molecules C6H12O6 = (0.0444 mol)(6.022 E23 molecules/mol)
⇒ molecules C6H12O6 = 2.674 E22 molecules
THE LIVING AND NON-LIVING THINGS ARE INTERDEPENDENT IN EACH OTHER............EVERY THINGS IN THE UNIVERSE IS IMPORTANT.....FOR EXAMPLE, THE RAW MATERIALS THAT ARE AVAILABLE ON THE SURFACE OF THE EARTH IS FINALLY CONVERTED INTPO THE USABLEPRODUCTS IR THE FINISHED GOODS.....THUS, THEY BOTH ARE EQUALLY IMPORTANT IN THIS UNIVERSE
