I. The solubility of NaCl at 25 degrees C would be between the solubilities at 20 and 30 degrees C. A reasonable answer would be 36 grams/100 g water
ii. From the table, it’s clear that the salts are more soluble at higher temperatures, indicating that an increase in temperature increases solubility.
iii. At 50 degrees C, a saturated ammonium chloride solution will have 50.6 grams of salt per 100 g water. At 20 degrees C, the solution can hold only 37.3 grams of salt per 100 g water. Thus, 13.3 grams of salt will precipitate per 100 grams of water.
Answer:
Explanation:
C + O2 → CO2
Mole of C = 24 g/(12 g/mole)
Mole of C = 2 mole
Mole of molecular O2 = 74 g/(32 g/mole)
Mole of molecular O2 = 2.3125 mole
Since mole of C < mole of O2, then C being the limiting reagent.
From the reaction, it shows that mole ratio between C and O2 = 1 : 1.
So, 2 moles of C will stoichiometrically react with 2 moles of O2 to generate 2 moles of CO2.
Avogadro's law states that :"equal volumes of all gases, at the same temperature and pressure, have the same number of molecules i.e. 6.02 x 10^23 molecules/mole.
Therefore, 2 moles of CO2 contain 2 moles x 6.02 x 10^23 molecules/mole = 1.204 x 10^24 molecules of CO2 is formed.
Answer:
30 moles
Explanation:
Water is H2O, meaning there is 2 Hydrogen atoms and 1 Oxygen atom. Oxygen is O2, because it is a diatomic molecule. (Hydrogen is also a diatomic molecule, so H2)
The equation, balanced, would have to be: 2H2 + O2 -----> 2H2O
I multiply 15 moles O2 by the molar ratio of (hydrogen/oxygen)
15 mol. O2 * (2 mol. H2/1 mol O2) = 30 moles of water
Answer:
Final volume 30.513 L.
Explanation:
According to general gas equation:
P₁V₁/T₁ = P₂V₂/T₂
Given data:
Initial volume = 17 L
Initial pressure = 2.3 atm
Initial temperature = 299 K
Final temperature = 350 K
Final volume = ?
Final pressure = 1.5 atm
Solution:
P₁V₁/T₁ = P₂V₂/T₂
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 2.3 atm × 17 L × 350 K / 299 K × 1.5 atm
V₂ = 13685 atm .L. K / 448.5 K . atm
V₂ = 30.513 L
