The hypothesis is that salt water freezes faster than fresh water.
The dependent variable is time taken for ice to appear.
The independent variable is presence or absence of salt
The constants are the amount of water in each tray, freezing conditions and length of time of exposure to freezing condition.
The control group is the tray to which salt was not added
The experimental group is the tray to which salt was added
The presence of solutes in a solution causes the freezing point depression.
A solution is made up of a solute and a solvent. In the presence of a solute, the freezing point of a pure solvent is decreased. This is because freezing point is a colligative property.
Colligative properties depend on the amount of solute present.
Hence, the pure water freezes faster (ice begin to appear earlier) than the salt water.
The hypothesis put forward in this experiment was found to be invalid by the experiment.
For more about colligative properties, see
brainly.com/question/10323760
Answer:
If we assume the molar volumes of water and ethanol 17.0 and 57.0 cm³/mol, respectively, Vmix = 20.5 cm³.
Explanation:
The molar volume of a substance is the ratio between the volume and the number of moles of the substance. It represents the volume that 1 mol of it occupies. Because we don't have access to page 24, let's assume the molar volumes of water and ethanol 17.0 and 57.0 cm³/mol, respectively.
The volume of mixture (Vmix) is the sum of the volume of each substance, which is the number of moles multiplied by molar volume, so:
Vmix = 0.300*57 + 0.200*17
Vmix = 17.1 + 3.4
Vmix = 20.5 cm³
CH3 is a methyl radical, which is formed by removing the hydrogen atom from methane, so the hybridization is SP^3
First, we need to get the molar mass of:
KClO3 = 39.1 + 35.5 + 3*16 = 122.6 g/mol
KCl =39.1 + 35.5 = 74.6 g/mol
O2 = 16*2 = 32 g/mol
From the given equation we can see that:
every 2 moles of KClO3 gives 3 moles of O2
when mass = moles * molar mass
∴ the mass of KClO3 = (2mol of KClO3*122.6g/mol) = 245.2 g
and the mass of O2 then = 3 mol * 32g/mol = 96 g
so, 245.2 g of KClO3 gives 96 g of O2
A) 2.72 g of KClO3:
when 245.2 KClO3 gives → 96 g O2
2.72 g KClO3 gives → X
X = 2.72 g KClO3 * 96 g O2/245.2 KClO3
= 1.06 g of O2
B) 0.361 g KClO3:
when 245.2 g KClO3 gives → 96 g O2
0.361 g KClO3 gives → X
∴ X = 0.361g KClO3 * 96 g / 245.2 g
= 0.141 g of O2
C) 83.6 Kg KClO3:
when 245.2 g KClO3 gives → 96 g O2
83.6 Kg KClO3 gives → X
∴X = 83.6 Kg* 96 g O2 /245.2 g KClO3
= 32.7 Kg of O2
D) 22.4 mg of KClO3:
when 245.2 g KClO3 gives → 96 g O2
22.4 mg KClO3 gives → X
∴X = 22.4 mg * 96 g O2 / 245.2 g KClO3
= 8.8 mg of O2
