Answer:
4.083 * 10^20 atoms.
Explanation:
One Mole of phosphorus contains 6.022 * 10^23 atoms (Avogadros number)'
Since 1 mole of Phosphorus has a mass of 30.974 grams, 21 milligrams has
6.022 * 10^23 * 0.021 / 30.974
= 0.004083 * 10^23
= 4.083 * 10^20
Answer:
1244 students
Explanation:
That would be y = 82*3 + 998
= 1244.
Particles of gas are more scarcely placed as compared to that of liquid.
the intermolecular forces will be less in gaseous state and hence is less stable
Answer:
2.2 °C/m
Explanation:
It seems the question is incomplete. However, this problem has been found in a web search, with values as follow:
" A certain substance X melts at a temperature of -9.9 °C. But if a 350 g sample of X is prepared with 31.8 g of urea (CH₄N₂O) dissolved in it, the sample is found to have a melting point of -13.2°C instead. Calculate the molal freezing point depression constant of X. Round your answer to 2 significant digits. "
So we use the formula for <em>freezing point depression</em>:
In this case, ΔTf = 13.2 - 9.9 = 3.3°C
m is the molality (moles solute/kg solvent)
- 350 g X ⇒ 350/1000 = 0.35 kg X
- 31.8 g Urea ÷ 60 g/mol = 0.53 mol Urea
Molality = 0.53 / 0.35 = 1.51 m
So now we have all the required data to <u>solve for Kf</u>:
