chemistry, physics, anyone?

hey anyone who think they're smart enough to take on uec sci!! knock yourselves out! =)
note: if cant view the pictures, open them. i drew it using paint!!! took such a long time!!! T-T

chemistry:
1. the solubility (g/100g water) of a substance at 30K is 25g. at the said temperature, 20g of the same substance is added into 50g of water, thoroughly stirred and then filtered. The mass percentage of the substance in the solutioin obtained is ______________ x 100%
A. 25/100+25 B. 20/50+20 C. 25/100 D. 20/50

ans: A
ques: are you sure? shouldnt it be like. B? smart ppl, please, explain to me WHY, if the ans is really A! =)

2. A saturated copper sulphate solution is prepared at a certain temperature. If 25g of this saturated solution is heated to dryness and 5g of copper sulphate crystal is obtained, what is the solubility of copper sulphate in 100g of water?
A. 25 B. 20 C. 5 D. 2

ans: A
ques: are you sure? i think its B. once again, smart ppl, please explain to me HUAI, if the ans is really A. =)

physics:
1.
An uncharged insulator conductor AB is placed near another positively charged insulated conductor C and both are connected to form a system as shown in this figure. What is the charge of B in each of the following cases? (select positvie charge, negative charge or uncharged)

1. K1 and K2 are both open

2. K1 is closed while K2 is open

3. K1 is opened while K2 is closed

4. K1 is first closed and then opened, C is finally removed and K2 is opened throughout.

ans: 1. positive 2. positive 3. uncharged 4. positive 5. negative

ques: i dont understand. when k1 is opened, shouldnt it be uncharged? the whole thing? in a normal situation, without earthing, A would be negatvie and B would be positive right? does that happen only when K1 is closed? cant it happen when K1 is open? and with earthing, should the negative go only? then the whole thing would be positive? HUH?? explain the concept of this thingy to me please!!!

2. An insulated metal rod carrying positive charges is placed between two uncharged and insulated metal spheres. This is the answer to the ques: which of the following figures show the possible charges on the metal spheres.

my ques: why is the 2nd charge of the right sphere uncharged? i thought when you earth, only the negative charges would go? theres another option (mcq) where the it is the same, just that the 2nd charge of the right sphere is positive. i just dont get it. earthing should be negative charges that go right? not positive?

3. A student holding a positively charged insulated metal rod approaches the metal cap of an uncharged electroscope, then taps his finger on the metal cap lightly, and finally moves away the positively charged metal rod. what will happen to the electroscope instantly?

A. uncharged B. positively charged C. negatively charged D. cannot be determined

ans: B

ques: why? id ont really understand the ques. explain the whole thing!

4.

The three resistors int eh circuit shown have equal resistance R. Find the effective resistance between x and y.
A. 1/3 R B. 2/3 R C. 3/2 R D. 3 R

ans: C

ques: what is effective resistance? how do you calculate that?

5.

The resistors R1 and R2 and R3 are connected to a cell of 3V in voltage. Given that R1=4, R2= 12 and the voltage acrsos R1 is 1.5V, then R3=?

A. 12 B. 8 C. 6 D. 4

ans: C
ques: voltage across R1 is where? and how come there can be 2 voltages in a circuit? i dont get why. dow do you count this thing?

6. Resistors R1 and R2 are connected in series to the terminals of a source. If the voltage between the terminalas of R1 is 1/10 that of the source, then the resistance of R2 is how many times that of R1?

A. 11 B. 10 C. 9 D. 8

ans: no ans given. -.-

ques: how do you actually count this???

T________T i have a headache! just a few more to go!

7.

1. The light bulbs are labeled as "6V, 4W", "6V, 6W' and "6V, 12W" respectively. Find the equivalent resistance of the circuit.

ans: L1=9; L2=6; L3=3; total= 9/2 ohm

2. If the total electric potential of the dry cells remains at 6V, find the current passing through L2 and L3

ans: V=IR; 6=I (9); I=2/3A; I2=I3=2/3A

ques: why cant i use the formula P=VI? huhhhh?

3. If the total electric potential of the dry cells remains at 6V, find the voltages at the terminals of L2 and L3

ans: V2=I2R2; =2/3*6; =4V V3=I3R3; =2/3*3; =2V

ques: i thought the voltage was arleady given??? 6V???

4. If the total electric potential of the dry cells remains at 6V, find the total electric energy used by the 3 bulbs in 1 minute.

ans: W=IVT; W=V*V/R *T; = 36/4.5 *60; =480J

ques: cant i use W=Pt? P=VI... or just W=IVT without sub-ing I=V/R in? and in cases like this, how do you know which I or which R to take? add all up?