Answer:
1) pure water
2) 0.75 m CaCl2
3) 1.0 m NaCl
4) 0.5 m KBr
5) 1.5 m glucose (C6H12O6)
Explanation:
Boiling point elevation is a colligative property. Coligative properties are properties that depend on the amount of solute present in the system. The boiling point of solvents increase due to the presence of solutes.
The boiling point elevation depends on the number of particles the solute forms in solution and the molality of the solute. The more the number of particles formed by the solute and the greater the molality of the solute, the greater the magnitude of boiling point elevation.
The order of decreasing hoping point elevation is;
1) 0.75 m CaCl2
2) 1.0 m NaCl
3) 0.5 m KBr
4) 1.5 m glucose (C6H12O6)
Answer:
6CO2 + 6H20 ----> C6H12O6 + 6O2
Explanation:
(Carbon Dioxide) + (Water) ------> (Glucose) + (Oxygen)
Answer:
A. fluorine, 1.79 moles
Explanation:
Given parameters:
Mass of carbon = 87.7g
Mass of fluorine gas = 136g
Unknown:
The limiting reactant and the maximum amount of moles of carbon tetrafluoride that can be produced = ?
Solution:
Equation of the reaction:
C + 2F₂ → CF₄
let us find the number of the moles the given species;
Number of moles =
C; molar mass = 12;
Number of moles =
= 7.31moles
F; molar mass = 2(19) = 38g/mol
Number of moles =
= 3.58moles
So;
From the give reaction:
1 mole of C requires 2 moles of F₂
7.31 moles of C will then require 2 x 7.31 moles of F₂ = 14.62moles
But we have 3.58 moles of the F₂;
Therefore, the reactant in short supply is F₂ and it is the limiting reactant;
So;
2 moles of F₂ will produce mole of CF₄
3.58 moles of F₂ will then produce
= 1.79moles of CF₄
Answer:
The mass of silicon in kilograms in Earth's crust is
.
Explanation:
Mass of Earth =
(1 ton= 2000 lb)
(1 lb =453.6 g)
1 ton = 2000 × 453.6 g =907,200 g
Mass of Earth =
Percentage of earth crust = 0.50%
Mass of earth crust = M


Percentage of the silicon in Earth's crust = 27.2 %
Mass of silicon in in Earth's crust = m



1000 g = 1 kg
The mass of silicon in kilograms in Earth's crust is
.
Water's specific heat capacity is 4200 J/Kg°C
95-28=67
72.5grams in kg is 0.0725kg
Energy = 67×0.0725×4200
Energy = 20,401.5 J or 20.4015 kJ