Answer:
D. The mixing of warm and cold water
Explanation:
Answer:
A. 85.6 g
= 0.0856 kg.
B. 0.00027 mol/g
= 0.27 mol/kg.
C. 8.39 %
Explanation:
Given:
Molar concentration = 0.25 M
Molar weight of sucrose = 342.296 g/mol
Density of solution = 1.02 g/mL
Mass of water = 934.4 g.
Density in g/l = 1.020 g/ml * 1000ml/1 l
= 1020 g/l
Mass of solution in 1 l of solution = 1020 g
Mass of solution = mass of solvent + mass of solute
Mass of sucrose = 1020 - 934.4
= 85.6 g of sucrose in 1 l of solution.
A.
Density of sucrose = mass/volume
= molar mass/molar concentration
= 342.296 * 0.25
= 85.6 g/l
Number of moles = mass/molar mass
= 85.6/342.296
= 0.25 mol
B.
Molality = number of moles of solute/mass of solvent
= 0.25/934.4
= 0.00027 mol/g
C.
% mass of sucrose = mass of sucrose/total mass of solution * 100
= 85.6/1020 * 100
= 8.39 %
Sika have more food choices because they eat both grasses and shrubs, compared to the white-tailed dear who only eats shrubs.
Answer:
6.564×10¹⁶ fg.
Explanation:
The following data were obtained from the question:
Mass of beaker = 76.9 g
Mass of beaker + salt = 142.54 g
Mass of salt in fg =?
Next, we shall determine the mass of the salt in grams (g). This can be obtained as follow:
Mass of beaker = 76.9 g
Mass of beaker + salt = 142.54 g
Mass of salt =?
Mass of salt = (Mass of beaker + salt) – (Mass of beaker)
Mass of salt = 142.54 – 76.9
Mass of salt = 65.64 g
Finally, we shall convert 65.64 g to femtograms (fg) as illustrated below:
Recall:
1 g = 1×10¹⁵ fg
Therefore,
65.64 g = 65.64 g × 1×10¹⁵ fg / 1g
65.64 g = 6.564×10¹⁶ fg
Therefore, the mass of the salt is 6.564×10¹⁶ fg.
0.781 moles
Explanation:
We begin by balancing the chemical equation;
O₂ (g) + 2H₂ (g) → 2H₂O (g)
21.8 Liters = 21.8 Kgs
To find how many moles are in 28.1 Kg H₂O;
Molar mass of H₂O = 18 g/mol
28.1/18
= 1.56 moles
The mole ratio between water vapor and oxygen is;
1 : 2
x : 1.56
2x = 1.56
x = 1.56 / 2
x = 0.781
0.781 moles
