a b c d no a no b yes c no d ok
This uses the concept of freezing point depression. When faced with this issue, we use the following equation:
ΔT = i·Kf·m
which translates in english to:
Change in freezing point = vant hoff factor * molal freezing point depression constant * molality of solution
Because the freezing point depression is a colligative property, it does not depend on the identity of the molecules, just the number of them.
Now, we know that molality will be constant, and Kf will be constant, so our only unknown is "i", or the van't hoff factor.
The van't hoff factor is the number of atoms that dissociate from each individual molecule. The higher the van't hoff factor, the more depressed the freezing point will be.
NaCl will dissociate into Na+ and Cl-, so it has i = 2
CaCl2 will dissociate into Ca2+ and 2 Cl-, so it has i = 3
AlBr3 will dissociate into Al3+ and 3 Br-, so it has i = 4
Therefore, AlBr3 will lower the freezing point of water the most.
Mass × velocity = momentum
74 x 15 = momentum
Answer:
you would find it in section 3 as these are semi-metals
Explanation:
Answer:
76.56g
Explanation:
Firstly, to do this we need a correct and balanced equation for the decomposition of potassium chlorate.
2KClO3 —-> 2KCl + 3O2
From the balanced equation, we can see that 2 moles of potassium chlorate yielded 3 moles of oxygen gas
We need to know the actual number of moles of oxygen gas produced. To do this, we divide the mass of the oxygen gas by its molar mass. Its molar mass is 32g/mol
The number of moles is thus 30/32 = 0.9375 moles
Now we can calculate the number of moles of potassium chlorate decomposed.
We simply do this by (0.9375 * 2)/3 = 0.625 moles
Now to get the number of grammes of potassium chlorate decomposed, we simply multiply this number of moles by the molecular mass. The molecular mass of KClO3 is 39 + 35.5 + 3(16) = 122.5g/mol
The amount in grammes is thus 122.5 * 0.625 = 76.56g
