Question

1.)30 cm^3 of water is added to 50cm^3 of a solution with molarity M at a constant temperaturee.after dilution the molarity of the solution is what?

i would say 3/5 M

2.)The number of molecules of sulphur dioxide in 5.6 dm^3 at stp is

a)1.5 x 10^23
b)5.6 x 10^23
c)6 x 10^23
d)2.4 x 10^24
e)3.6 x 10^24

I WOULD SAY C
3.)What mass of oxygen is in 34.5g of Al2O3 .2H2o
a)20g
b)12g
c)16g
d)6.9g
e)Not possible to determine from info given

i would say E

4.)What mass of iron would contain the same number of atoms as 10g of silicon
a)2.8g'
b)5.0g
c)5.6g
d)10g'
e)20g

i think its c

Answer 1

@JoannaBlackwelder can u check and c if i am correct please.

Answer 2

I don't get 3/5M for the first question. 1 cm^3 = 1 mL, so we have 0.05 L of M molarity solution. This would be equivalent to 0.05*M moles. If we added 0.03 L to the total volume, but no more moles, the volume would be 0.08 L. The molarity of the solution would therefore be 0.05M/0.08 which reduces to 5/8M.

Answer 3

ohhh ok i c what u did

Answer 4

Cool! I also don't get C for the second question. I used the conversion factors of 1 dm^3=1L, 22.4 L of gas at STP =1mol and 6.022x10^23=1mol. Take a look at that one again and tell me how many mols of SO2 you get.

Answer 5

You can calculate the mass of Oxygen in 34.5 g Al2O3.2H2O by using the molar mass of Al2O3.2H2O, the molar ratio between Al2O3.2H2O, and the molar mass of Oxygen.

Answer 6

*the molar ratio between Al2O3.2H2O and Oxygen

Answer 7

e

Answer 8

sorry 4 the long pause was tryna work it out

Answer 9

No problem at all. I am assuming you mean e for problem 2?

Answer 10

yes

Answer 11

How did you get that?

Answer 12

oh its A i did 5.6/22.4=0.25

0.25x6x10^23

Answer 13

=1.5x10^23

Answer 14

Yes! Excellent!

Answer 15

:D glad i got it

Answer 16

Me too! Did you look at the third question again?

Answer 17

oh its wrong let me c

Answer 18

I commented on it a little earlier.

Answer 19

oh so it is right then

Answer 20

No, the answer for the third question is not e.

Answer 21

i dont know how to get this one i tried multiplying the grams to get it

Answer 22

You can only relate the masses by using a mole ratio. Take 34.5 g of Al2.2H2O *(1 mol Al2O3.2H2O/137.992 g Al2O3.2H2O)*(5 mol Oxygen/1 mol Al2O3.2 H2O) *(15.9994 g Oxygen/1 mol Oxygen) = 20.0 g Oxygen

Answer 23

wow i wouldnt have gotten that one

Answer 24

Yeah, those kind take some practice. By the way, there are 5 moles of Oxygen for every 1 mole of the molecule because there are 5 atoms of oxygen in each molecule.

Answer 25

ok

Answer 26

The 4th problem is somewhat similar to the third. Try taking another look at it.

Answer 27

lol is it e?

Answer 28

Yes! :) How did you get it?

Answer 29

using the mole ratio

Answer 30

Alright! I think you are catching on! :) Keep practicing!

Answer 31

Joanna thank you so much for your help i REALLY !appreciate it thanks so much

Answer 32

You're welcome. :)

Answer 33

enjoy the rest of your day bye!

Answer 34

Thanks, you too!