2. Avogadro's constant is 6.02*10^23
3,4, and 5 are ok.
I'll check the others asap.
I was just wondering if someone would be able to critique my answers for the following Chem questions? they range from pretty basic to a bit more...
2)which of the following is the definition of the value of Avogadro's constant:
b) the # of Carbon atoms in exactly 12g of carbon-12?
3)the mass % of Ca in milk is 0.118%. How many Ca atoms are present in 250.0g of milk?
b) 4.43*10^23 [250gmilk*0.118gCa/40.078g.molCa*6.0221*10^23?]
4)What is the mass of 7.1*10^2moles of tin(Sn)?
a)8.4*10^4g
[7.1mol*10^2 * 118.710g/mol=8.4*10^4g]
5)An element has 2 naturally occurring isotopes. The first has a mass of 78.9183u & an abundance of 50.54%, while the second has a mass of 80.9163u. The atomic mass (weighted)of the element is therefore:
b) 79.91u
[49.46*80.9163=40.0212 + 50.54%*78.9183=39.8853 = 79.91u
7)A 200.0g sample of pure MgCl2 contains:
b)2.101 formula units of MgCl2
[200gMgCl2*1mol/95.211g/mol MgCl2 =2.101 formula units
???10)How much water is present in 83.0g of Li3PO4~12H2O?
a=0.250mol, b=0.717mol, c=2.61mol, d=3.00mol...I got "D"=3.00 mol
17)WHich of the following solutions of electrolytes are expected to undergo complete ionization?
a)0.1M LiCl, c)0.1M HCl e)both a&c...I chose "E'both A&C"
I don't understand how to do the following questions:
????14)---Given the reaction:
4Zn + K2Cr2O7 + 7H2SO4 -> 2CrSO4 + K2SO4 + 7H2O
How many moles of CrSO4 are produced by allowing 9moles of Zn, 3moles of K2Cr2O7 and 14moles of H2SO4 to react?
answers:
a)1mol, b)2mol, c)3mol, d)4mol, e)5mol
????16)What volume of 0.220M HCl is required to exactly neutralize 10.0mL of 0.085M NaOH?
a)7.7mL, b)1.0mL, c)0.39mL, d)3.9mL, e)0.19mL
????25) A solution of unknown acid is made by dissolving 0.0724g of this acid in water to make a 20mL solution. This unknown acid is known to react with NaOH in a 1:1mole ratio. The solution of this acid is titrated with 17.20mL of 0.1152M NaOH to reach equivalency. What is the identity of the unknown acid????
a)HF, b)HCl, c)HBr, d)CH3COOH, e)none of the above...of course it "isn't 'd"
but how do I work this out to find out, even if it is "e"none of the above???
Thank you for your time...it's greatly appreciated!!!
9 answers
It would be better to put just one (perhaps two) questions per post.
7 is ok.
10 is ok.
17. Looks ok but I don't know what your other choices were. Solutions of both LiCl and HCl will be 100 ionized.
7 is ok.
10 is ok.
17. Looks ok but I don't know what your other choices were. Solutions of both LiCl and HCl will be 100 ionized.
????14)---Given the reaction:
4Zn + K2Cr2O7 + 7H2SO4 -> 2CrSO4 + K2SO4 + 7H2O
How many moles of CrSO4 are produced by allowing 9moles of Zn, 3moles of K2Cr2O7 and 14moles of H2SO4 to react?
answers:
a)1mol, b)2mol, c)3mol, d)4mol, e)5mol
Just convert mols of each of the materials listed to mols CrSO4. The smallest number will be the correct answer. This is just another example ;of a limiting reagent.
9 mols Zn x (2 mols CrSO4/4 mols Zn) =xx
3 mols K2Cr2O7 x (2 mols CrSO4/1 mol K2Cr2O7) = ??
14 mols H2SO4 x (2 mols CrSO4/7 mols H2SO4) = zz
????16)What volume of 0.220M HCl is required to exactly neutralize 10.0mL of 0.085M NaOH?
a)7.7mL, b)1.0mL, c)0.39mL, d)3.9mL, e)0.19mL
HCl + NaOH ==> NaCl + H2O
mols NaOH = L x M = 0.0100 x 0.085 = ??
mols HCl must be equal to mols NaOH (from the equation it is 1:1)
Mols HCl = L x M. You have mols and M. Solve for L.
????25) A solution of unknown acid is made by dissolving 0.0724g of this acid in water to make a 20mL solution. This unknown acid is known to react with NaOH in a 1:1mole ratio. The solution of this acid is titrated with 17.20mL of 0.1152M NaOH to reach equivalency. What is the identity of the unknown acid????
mols NaOH = L x M = 0.01720 L x 0.1152 M = ??
mols NaOH = mols acid from the problem.
mols acid = grams/molar mass.
You know mols acid and grams. Solve for molar mass and that should give you a clue as to the identity of the acid. If not, calculate the molar of the compounds in each of the answers and compare. Your compound is there.
a)HF, b)HCl, c)HBr, d)CH3COOH, e)none of the above...of course it "isn't 'd"
but how do I work this out to find out, even if it is "e"none of the above???
Check my work carefully.
4Zn + K2Cr2O7 + 7H2SO4 -> 2CrSO4 + K2SO4 + 7H2O
How many moles of CrSO4 are produced by allowing 9moles of Zn, 3moles of K2Cr2O7 and 14moles of H2SO4 to react?
answers:
a)1mol, b)2mol, c)3mol, d)4mol, e)5mol
Just convert mols of each of the materials listed to mols CrSO4. The smallest number will be the correct answer. This is just another example ;of a limiting reagent.
9 mols Zn x (2 mols CrSO4/4 mols Zn) =xx
3 mols K2Cr2O7 x (2 mols CrSO4/1 mol K2Cr2O7) = ??
14 mols H2SO4 x (2 mols CrSO4/7 mols H2SO4) = zz
????16)What volume of 0.220M HCl is required to exactly neutralize 10.0mL of 0.085M NaOH?
a)7.7mL, b)1.0mL, c)0.39mL, d)3.9mL, e)0.19mL
HCl + NaOH ==> NaCl + H2O
mols NaOH = L x M = 0.0100 x 0.085 = ??
mols HCl must be equal to mols NaOH (from the equation it is 1:1)
Mols HCl = L x M. You have mols and M. Solve for L.
????25) A solution of unknown acid is made by dissolving 0.0724g of this acid in water to make a 20mL solution. This unknown acid is known to react with NaOH in a 1:1mole ratio. The solution of this acid is titrated with 17.20mL of 0.1152M NaOH to reach equivalency. What is the identity of the unknown acid????
mols NaOH = L x M = 0.01720 L x 0.1152 M = ??
mols NaOH = mols acid from the problem.
mols acid = grams/molar mass.
You know mols acid and grams. Solve for molar mass and that should give you a clue as to the identity of the acid. If not, calculate the molar of the compounds in each of the answers and compare. Your compound is there.
a)HF, b)HCl, c)HBr, d)CH3COOH, e)none of the above...of course it "isn't 'd"
but how do I work this out to find out, even if it is "e"none of the above???
Check my work carefully.
so for #14 I got: 4mols
#16: 1.0mL
#17...the other two choices were b-0.1MCH3COOH & then d) both A&B, that's why I didn't bother putting them in, considering b is not an electrolyte.
#25: b- HCl?
#2, about Av's #...the options were:
a)the # of Hydrogen atom in exactly 1g of hydrogen-1
b)the # of carbon atoms in exactly 12g of carbon-12
c)the # of carbon atom in exactly 1g of carbon-12
d)the # of hydrogen atoms in exactly 12g of hydrogen-12
e)the # of moles in exactly 1g of carbon-12.
thank you for your time again.
#16: 1.0mL
#17...the other two choices were b-0.1MCH3COOH & then d) both A&B, that's why I didn't bother putting them in, considering b is not an electrolyte.
#25: b- HCl?
#2, about Av's #...the options were:
a)the # of Hydrogen atom in exactly 1g of hydrogen-1
b)the # of carbon atoms in exactly 12g of carbon-12
c)the # of carbon atom in exactly 1g of carbon-12
d)the # of hydrogen atoms in exactly 12g of hydrogen-12
e)the # of moles in exactly 1g of carbon-12.
thank you for your time again.
14 is ok.
16. Check you math. I get
0.01 x 0.085 = 0.00085 = mols NaOH
mols HCl = L x M
0.00085 = L x 0.220 = 0.00386 L or 3.86 mL and rounded to two places is 3.8 mL.
17 was ok and still is. You are correct that CH3COOH is a weak electrolyte.
25 is HCl. Correct.
b is correct and that is what you had. I just didn't equate it since I had the 6.02*10^23 in mind.
16. Check you math. I get
0.01 x 0.085 = 0.00085 = mols NaOH
mols HCl = L x M
0.00085 = L x 0.220 = 0.00386 L or 3.86 mL and rounded to two places is 3.8 mL.
17 was ok and still is. You are correct that CH3COOH is a weak electrolyte.
25 is HCl. Correct.
b is correct and that is what you had. I just didn't equate it since I had the 6.02*10^23 in mind.
I made a typo above. The 3.86 rounded to two places is, of course, 3.9 mL and not 3.8 mL.
Thank you muchly!! (the questions are just from an old MT that we were to go over to study for the real MT this coming Sat.)
good luck.
thank you!
2 answers