Given parameters:
Mass of sucrose = 5g
Density of sucrose = 1.12g/mL
Percentage of sucrose per liter of cane juice = 12%
Unknown:
Volume of cane juice needed = ?
We need to establish the volume - density relationship. Density is the mass of a substance per unit volume.
Mathematically;
Density = 
Now solve for the volume of sucrose;
1.12g/mL = 
Volume = 
= 4.46mL = 4.46 x 10⁻³L since 1000mL = 1L
Since 12% of 1 liter of cane juice is sucrose;
12% of x liter of cane juice = 4.46 x 10⁻³L
Volume of cane juice = 4.46 x 10⁻³ x 
= 0.037L
Volume of cane juice is 0.037L
Answer:
z = 18,46%
Explanation:
First we need to discover the mole number of the O2 produced so we use PV=nRT
734 torr * 57.2*10^-3 L = n 62.3637 * 295.15 K
n = 
n = 
n = 2,28 * 10^-3 mole
The reaction to the decomposition is:
2 NaClO3 --> 2 NaCl + 3 O2
2 mole NaClO3 - 3 moles O2
x - 2,28*10^-3 moles O2
X= 1,52*10^-3 moles of NaClO3
1 moles of NaClO3 - 106,44 g
1,52*10^-3 moles of NaClO3 - y
y = 0,1617 g
0,8765 g - 100%
0,1617 g - z
z = 18,46%
The answer is b.low frequency,high energy
Answer: - 1.86°C
Explanation:
The depression of freezing points of solutions is a colligative property.
That means that the depression of freezing points of solutions depends on the number of molecules or particles dissolved and not the nature of the solute.
To solve the problem follow these steps:
Data:
Tf = ?
solute = glucosa (this implies i factor is 1)
mass of solue = 36.0 g
mass of water = 500 g
Kf = 1.86 °/m
mm glucose = 180.0 g / mol
2) Formulas
Tf = Normal Tf - ΔTf
ΔTf = i * kf * m
m = number of moles of solute / kg of solvent
number of moles of solute = mass in grams / molar mass
3) Solution
number of moles of solute = 36.0 g / 180.0 g/mol = 0.2 mol
m = 0.2 mol / 0.5 kg = 1.0 m
ΔTf = i * Kb * m = 1 * 1.86 °C/m * 1 m = 1.86°C
Tf = 0°C - 1.86°C = - 1.86°C
Answer: - 1.86 °C
