Some chemistry help please :\

[whc235]

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1. Consider a 400 mL of a solution of containing 0.20 M NH3 and 0.30 M NH4Cl to which 100 mL of 0.05 M NaOH is added. What is the resulting pH after the addition of the NaOH?: *


3. If 0.5423 g of glucose (C6H12O6) is combusted, what mass, in grams, of CO2 will be produced?: *


7. Balance the following equation by the half-reaction method for the pH condition specified:

SO32- (aq) + MnO4- (aq) → SO42- (aq) + Mn2+(aq) (acidic): *

8. Dianabol is one of the anabolic steroids that has been used by some athletes to increase the size and strength of their muscles. It is similar to the male hormone testosterone. Some studies indicate that the desired effects of the drug are minimal, and the side effects, which include sterility and increased risk of liver cancer and heart disease, keep most people from using it. The molecular formula of Dianabol, which consists of carbon, hydrogen, and oxygen, can be determined using the data from two different experiments. In the first experiment, 14.765 g of Dianabol is burned, and 43.257 g CO2 and 12.395 g H2O are formed. In the second experiment, the molecular mass of Dianabol is found to be 300.44. What is the molecular formula for Dianabol?: *
a. C10H14O
b. C20H28O2
c. C30H42O9
d. C16H28O5

Thanks guys.. my chemistry is kinda rusty :\

 

[UndergradGuy7]

#1 I think is a buffer problem use pH = pKa + log (A-/HA) or pH = pKa + log (B/BH+) (same thing just acid or base)

so normally you would set it up as pH = pKa(of NH4+) + log (0.20M NH3/0.30M NH4+) but here, we also add NaOH. NaOH is a base and will react with the acid (NH4+) to make it into NH3.

So now instead of having 0.08 moles of NH3 (0.20M NH3 * 0.400 L), we also form some more because of the NH4+ + NaOH to NH3 reaction. We form 0.1 L NaOH * 0.05 M NaOH = 0.005 moles of NH3 extra. So we add them together 0.08 + 0.005 moles = 0.085 moles NH3.

Now do the same for the NH4+ except, we are losing NH4+ instead of gaining it. So originally we had 0.3M NH4+ * 0.4L = 0.12 moles and we lost 0.005 moles (same amount of moles as above from the NH3 being formed). So 0.12 moles - 0.005 = 0.115 moles NH4+

Plug it into the equation and find the pKa value from the back of your chemistry book/or it is listed in the problem.


The overall simplified setup is pH=pKa + log [(0.2*0.4)+(0.1*0.05)] / [(0.3*0.4)-(0.1*0.5)]



#3 The reaction is a combustion so C6H12O6 + O2 -> CO2 + H2O. We need to balance it so... C6H12O6 + 6O2 -> 6CO2 + 6H2O
Convert grams of glucose to moles. Then use the mol ratios to convert to moles of CO2 and finally grams of CO2.
0.5423 g/(MW of glucose) * (6 mol CO2/ 1 mol glucose) * (MW of CO2 / 1 mol CO2) = answer

#7
SO32- (aq) + MnO4- (aq) → SO42- (aq) + Mn2+(aq) (acidic):
First find the 2 half reactions. To do this find which is being oxidized an which is reduced. S in SO32- is +4 but then becomes +6 in SO42-. It is oxidized. Mn in MnO4- is +7. Mn in the products is 2+. It is reduced.

So
First we balance other atoms in the formulas if we had any other than H+ or oxygen. Or if the amounts of S were different. But here we do not.
SO32- -> SO42-
balance O's with H2O.
SO32- + H2O -> SO42-
Balance H+'s with H+.
So SO32- + H2O -> SO42- + 2H+.
Balance charges by adding electrons.
Left side has SO3 2- so left side is only 2-. Right is is SO42- and 2H+ so it is -2 + 2 = 0 charge. The right side needs 2 electrons.
SO32- + H2O -> SO42- + 2H+ + 2e-

Do the same for the MnO4- -> Mn2+ and add the SO32- equation SO32- + H2O -> SO42- + 2H+ + 2e- with the MnO4- you get and cancel like terms.



Correct me if I am wrong. I think everything is right, but didn't re-check everything I might have mis-typed something.

 

[Chemboy23]

#3. C6H12O6 + O2 ------> 6CO2 + 6H2O

This is a stochiometry problem....just take grams of glucose to moles of glucose/mole of CO2 and then go from moles of CO2 to grams of CO2...I'll try below lol.

0.5423g C6H12O6 x 1 mol C6H12O6 x 6 mol CO2 x 44.01g CO = ~0.7948gCO2
180.18g C6H12O6 1 mol C6H12O6 1 mol CO2

 

[myheart01]

8. Dianabol is one of the anabolic steroids that has been used by some athletes to increase the size and strength of their muscles. It is similar to the male hormone testosterone. Some studies indicate that the desired effects of the drug are minimal, and the side effects, which include sterility and increased risk of liver cancer and heart disease, keep most people from using it. The molecular formula of Dianabol, which consists of carbon, hydrogen, and oxygen, can be determined using the data from two different experiments. In the first experiment, 14.765 g of Dianabol is burned, and 43.257 g CO2 and 12.395 g H2O are formed. In the second experiment, the molecular mass of Dianabol is found to be 300.44. What is the molecular formula for Dianabol?: *
a. C10H14O
b. C20H28O2
c. C30H42O9
d. C16H28O5

Answer is B
First, find the molecular mass
a.C10H14O=12*10+14+16=150g/mol
b.C20H28O2=12*20+28+16*2=300g/mol
c.C30H42O9=12*30+42+16*9=528g/mol
d.C16H28O5=12*16+28+16*5=300g/mol
therefore, answer is b or d

1. in case of b
C20H28O2 + O2 = 20CO2 + 14H2O
O2: excess substance --> never mind
14.8g of C20H28O2 = (14.8/300)=0.0493 mol of C20H28O2
g of CO2 = 0.0493*20*(12+16*2)= 43.38g --> match with question
g of H2O = 0.0493*14*(2+16)=12.42g --> match with question
therefore this is correct

2. in case of d
C16H28O5 + O2 = 16CO2 + 14H2O
14.8g of C20H28O2 = (14.8/300)=0.0493 mol of C20H28O2
g of CO2 = 0.0493*16*(12+16*2)= 34.71g --> not match with question
g of H2O = 0.0493*14*(2+16)=12.42g --> match with question
therefore this is incorrect

 

[prsndwg]

Can someone explain number 1 to me please.
also, for number 7, is the answer C?

 

[prsndwg]

:-?

 

[UndergradGuy7]

Can someone explain number 1 to me please.
also, for number 7, is the answer C?

I thought I explained number 1 pretty good above. I don't know if you missed it or didn't understand it. Check above.

I got C for number 7 too.

 

[hausee]

Ok for 7 I do this way:

SO3 2- => SO4 2- oxidation number on S change from +4 => +6 so lost 2e
MnO4 - => Mn 2+ oxidation number on Mn change from +7 => +2 so gain 5e

in order to let gain and lost e balance, All Mn will have 2 in front of them and all S will have 5 in front of them. Put them together and count how many O on left and right, you will see right need 3 more O, so add 3 H2O. finally left need 6 H+ to balance 6H we added on the right through H2O.

choice is C.