Answer:
A. percentage mass of iron = 5.17%
percentage mass of sand = 8.62%
percentage mass of water = 86.205%
B. (Iron + sand + water) -------> ( iron + sand) ------> sand
C. The step of separation of iron and sand
Explanation:
A. Percentage mass of the mixtures:
Total mass of mixture = (15.0 + 25.0 + 250.0) g =290.0 g
percentage mass of iron = 15/290 * 100% = 5.17%
percentage mass of sand = 25/290 * 100% = 8.62%
percentage mass of water = 250/290 * 100% = 86.205%
B. Flow chart of separation procedure
(Iron + sand + water) -------> separation by filtration using filter paper and funnel to remove water --------> ( iron + sand) -----------> separation using magnet to remove iron ------> sand
C. The step of separation of iron and sand by magnetization of iron will have the highest amount of error because during the process, some iron particles may not readily be attracted to the magnet as they may have become interlaced in-between sand grains. Also, some sand particle may also be attracted to the magnet as they are are borne on iron particles.
Answer:
Sunlight
Water
And carbon dioxide
Additional information :
6CO2 + 6H2O → C6H12O6 + 6O2
<u>Answer:</u> Group 1 ions are known as cations and Group 17 ions are known as anions.
<u>Explanation:</u>
Ions are formed when an atom looses or gains electrons.
If an atom gains electrons, it leads to the formation of negative ions known as anions. <u>For Example:</u> Fluorine is a Group 17 element which gains 1 electron to form 
ions.
If an atom looses electrons, it leads to the formation of positive ions known as cations. <u>For Example:</u> Sodium is a Group 1 element which looses 1 electron to form 
ions.
Hence, group 1 ions are known as cations and Group 17 ions are known as anions.
Answer:
Molar mass of solute is 89.28 g/m
Explanation:
Colligative property of freezing point depression to solve this:
ΔT = Kf . m . i
i = number of particles, dissolved in solution. In this case, it is a nonelectrolyte, so i = 1.
m = molalilty (mol of solute/1kg of solvent
ΔT = T° freeze pure solvent - T° freeze solution
0°C - (-2.50°C) = 1.86 °C/m . m
2.50°C / 1.86 m/°C = m
1.34 mol solute/kg solvent = m
This means, that in 1000 g of solvent, we have 1.34 moles but we have 250 g of solvent, so let's make a rule of three.
1000 g ____ 1.34 moles
250 g _____(2.50 . 1.34) / 1000 = 0.336 moles
To find the molar mass, we divide mass / moles
30 g/ 0.336 moles = 89.28 g/m
