chemistry questions

[envy]

Can you guys help me out..
HOW MANY GRAMS OF CALCIUM CHLORIDE ARE NEEDED TO PREPARE 72 GRAMS OF SILVER CHLORIDE ACORDING TO THE FOLLOWING EUATION?

CaCl2(aq) + 2AgNO3(aq) Ca(NO3)2 (aq) + 2AgCl (s)

27 g
28g
29g
30g
none of the above

HOW MANY ml OF WATER MUST BE ADDED TO 65ml OF A 5.5 m SOLUTION OF NaOH IN ORDER TO PREPARE A 1.2-M NaOH SOLUTION

230ml
235ml
229 ml
1L
NONE OF THE ABOVE

IN THE REACTION:
N2 + 2O2 2 NO2

WHAT VOLUME OF NO2 IS PRODUCED FROM 7g OF NITROGEN GAS AT 27 DEGREES CELCIUS AND 0.9 atm?

13.7 L
13.7 ml
1.37 L
1.37 ml
137 L
WHAT IS THE MOLE FRACTION OF ETHYL ALCOHOL, C2H5OH, AND WATER, RESPECTIVELY, IN A SOLUTION MADE BY DISSOLVING 9.2g OF ALCOHOL IN 19g OF WATER. (MW H20 =18, MW C2H5OH=46)?

0.167, 0.833
0.833, 0.167
0.133, 0.877
0.67, 0.33
NONE OF THE ABOVE

AT 90 DEGREES CELCIUS, BENZENE HAS A VAPOR PRESSURE OF 1022 Torr, AND TOLUENE HAS A VAPOR PRESSURE OF 406 Torr. GIVEN THIS INFORMATION, WHAT IS THE COMPOSITION OF THE BENZENE-TOLUENE SOLUTION THAT WILL BOIL AT 1 atm PRESSURE AND 90 DEGREES CELCIUS, ASSUMING THAT THE SOLUTION IS IDEAL?

THE MOLE FRACTION OF BENZENE IS 0.574
THE MOLE FRACTION OF TOLUENE IS 0.574
THE MOLE FRACTION OF BENZENE IS 0.426
THE MOLE FRACTION OF TOLUENE IS 0.326
MORE THAN ONE OF THE ABOVE

WHAT IS THE MAXIMUM NUMBER OF ELECTRONS THAT CAN BE PRESENT IN AN ATOM WITH A PRINCIPAL QUANTUM NUMBER OF 5 (NOTE: THE FORMULA IS: 2N^2)?

100
50
25
20
10
WHAT IS THE NORMALITY OF A 2.0M SOLUTION OF PHOSPHORIC ACID, H3PO4, FOR AN ACID-BASE TITRATION?

0.67
2
3
6
5

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[dat_student]

Can you guys help me out..
HOW MANY GRAMS OF CALCIUM CHLORIDE ARE NEEDED TO PREPARE 72 GRAMS OF SILVER CHLORIDE ACORDING TO THE FOLLOWING EUATION?

CaCl2(aq) + 2AgNO3(aq) Ca(NO3)2 (aq) + 2AgCl (s)

27 g
28g
29g
30g
none of the above

Bad questions & bad answer choices. I can't approximate. looooool
I solve the 1st one for now. Hope others will help as well.


72 g / (107.87 + 35.5) g/mole = 0.502197 mole

0.502197 * 1 mole CaCl2 / 2 moles AgCl * (40.08 + 2 * 35.45) g / 1 mole CaCl2 = 27.8669 I guess this means 28g is the best answer looooooooooooool

P.S. If you don't use 10 sig figs you'll end up with 27.5 loooooooool

 

[dat_student]

....
WHAT IS THE MAXIMUM NUMBER OF ELECTRONS THAT CAN BE PRESENT IN AN ATOM WITH A PRINCIPAL QUANTUM NUMBER OF 5 (NOTE: THE FORMULA IS: 2N^2)?

100
50
25
20
10

You're given the formula: 2 * N ^ 2

2 * (5 ^ 2) = 2 * 25 = 50

WHAT IS THE NORMALITY OF A 2.0M SOLUTION OF PHOSPHORIC ACID, H3PO4, FOR AN ACID-BASE TITRATION?

0.67
2
3
6
5

2 * 3 (because H3PO4 has 3 H+s) = 6

 

[dat_student]

...
AT 90 DEGREES CELCIUS, BENZENE HAS A VAPOR PRESSURE OF 1022 Torr, AND TOLUENE HAS A VAPOR PRESSURE OF 406 Torr. GIVEN THIS INFORMATION, WHAT IS THE COMPOSITION OF THE BENZENE-TOLUENE SOLUTION THAT WILL BOIL AT 1 atm PRESSURE AND 90 DEGREES CELCIUS, ASSUMING THAT THE SOLUTION IS IDEAL?

THE MOLE FRACTION OF BENZENE IS 0.574
THE MOLE FRACTION OF TOLUENE IS 0.574
THE MOLE FRACTION OF BENZENE IS 0.426
THE MOLE FRACTION OF TOLUENE IS 0.326
MORE THAN ONE OF THE ABOVE
...

Benzene & Tolune refer to mole fractions

1) Benzene + Toluene = 1
2) 1022 * Benzene + 406 * Toluene = 760 Torr (i.e. 1 atm)
1 & 2 -> 1022 (1 - Toluene) + 406 (Toluene) = 760
(1022 - 406) Toluene = 1022 - 760
***********************
*Toluene = 262 / 616 = 0.425*
***********************
Benzene = 1 - Toluene = 1 - 0.425 = 0.575
***********************
*Benzene = 0.575*********
***********************
Thus, the correct answer is "THE MOLE FRACTION OF BENZENE IS 0.574"

 

[dat_student]

...
WHAT IS THE MOLE FRACTION OF ETHYL ALCOHOL, C2H5OH, AND WATER, RESPECTIVELY, IN A SOLUTION MADE BY DISSOLVING 9.2g OF ALCOHOL IN 19g OF WATER. (MW H20 =18, MW C2H5OH=46)?

0.167, 0.833
0.833, 0.167
0.133, 0.877
0.67, 0.33
NONE OF THE ABOVE
...

C2H5OH:
Mole = 9.2 g / 46 g/mole = 0.2

H20:
Mole = 19 g / 18 g/mole = 1.05

Mole Fraction of C2H5OH = 0.2 / ( 0.2 + 1.05) = 0.16 approx
Mole Fraction of H20 = 1 - 0.16 = 0.84 approx

answer = 0.167, 0.833

 

[envy]

C2H5OH:
Mole = 9.2 g / 46 g/mole = 0.2

H20:
Mole = 19 g / 18 g/mole = 1.05

Mole Fraction of C2H5OH = 0.2 / ( 0.2 + 1.05) = 0.16 approx
Mole Fraction of H20 = 1 - 0.16 = 0.84 approx

answer = 0.167, 0.833

Thank you.

 

[super marz]

wow...

you guys are.......smart

i need to start studying 4 dat....crap!