Answer:
2.3 * 10^-5
Explanation:
Recall that the solubility of a solute is the amount of solute that dissolves in 1 dm^3 or 1000cm^3 of solution.
Hence;
Amount of calcium oxalate = 154 * 10^-3/128.097 g/mol = 1.2 * 10^-3 mols
From the question;
1.2 * 10^-3 mols dissolves in 250 mL
x moles dissolves in 1000mL
x = 1.2 * 10^-3 mols * 1000/250
x= 4.8 * 10^-3 moldm^-3
CaC2O4(s) ------->Ca^2+(aq) + C2O4^2-(aq)
Hence Ksp = [Ca^2+] [C2O4^2-]
Where;
[Ca^2+] = [C2O4^2-] = 4.8 * 10^-3 moldm^-3
Ksp = (4.8 * 10^-3)^2
Ksp = 2.3 * 10^-5
Elements of Group 1 and group 2 in the periodic
table contain elements so reactive that they are never found in the free state
<u>Explanation</u>:
The metals in group 1 of periodic table consisting of 'alkali metals' which include lithium, potassium, sodium, rubidium, Francium and caesium. They are highly reactive because they have low ionisation energy and larger radius. The group 2 metals consist of 'alkaline earth metals' which include calcium, strontium, barium, beryllium, radium and magnesium. These alkaline earth metal have +2 oxidation number, hence are highly reactive.
These both group metals are mostly reactive and so are never found in a free state. When they are exposed to air they would immediately react with oxygen. Hence, are stored in oils to avoid oxidation.
Answer:
Work done, W = 128 kJ
Explanation:
Given that,
Weight of a mountain climber, F = 800 N
It climbs to a cliff that is 160 m high.
We need to find the work done by the mountain climber. The work done by an object is given by the formula as follows :
W = Fd
Put the values of F and d.
W = 800 N × 160 m
W = 128000 J
or
W = 128 kJ
So, 128 kJ of work is done by the mountain climber.
Answer:
True
Explanation:
Here is an example: chemical properties include flammability, toxicity, acidity, reactivity. we observe the changes of these properties. Therefore, It's true.
Answer:
We'll have 82 moles ZnO and 41 moles S
Explanation:
Step 1: data given
Number of moles Zinc (Zn) = 82 moles
Number of moles sulfur oxide (SO2) = 42 moles
Step 2: The balanced equation
2Zn + SO2 → 2ZnO + S
Step 3: Calculate the limiting reactant
For 2 moles Zinc we need 1 mol sulfur oxide to produce 2 moles zinc oxide and 1 mol sulfur
Zinc is the limiting reactant. It will completely be consume (82 moles). Sulfur oxide is in excess. There will react 82/2 = 41 moles
There will remain 42-41 = 1 mol SO2
Step 4: Calculate moles of products
For 2 moles Zinc we need 1 mol sulfur oxide to produce 2 moles zinc oxide and 1 mol sulfur
For 82 moles Zinc we'll have 82 moles of Zinc Oxide (ZnO)
For 82 moles Zinc we'll have 82/2 = 41 moles of sulfur
We'll have 82 moles ZnO and 41 moles S
