candie

Orangeburg, SC

#42 Aug 24, 2009
compare the composition of sucrose purified from sugar cane with the composition of sucrose purified from sugar beets
Pokay1kaduB

Bolivar, OH

#43 Aug 25, 2009
candie wrote:
compare the composition of sucrose purified from sugar cane with the composition of sucrose purified from sugar beets
Read the last page starting on page 26, we talked about the same exact question
Danny

New York, NY

#44 Aug 31, 2009
Good morning, I have to take this answer to my teacher tomorrow morning and I don't have the answer. please help

Four different containers are labeled C + 02, CO,CO2 and CO. based on the labels clasiffy, each as an element, a compound, a homogeneous mixture, or a heterogeneous mixture.Explain your reasoning.
Danny

New York, NY

#45 Aug 31, 2009
How could you decide whether a ring was 24 karat gold or 14 karat gold without damaging the ring?

thank you very much for your help
Keith

Miami, FL

#46 Sep 1, 2009
Danny: This is a FAMOUS 2,000 year old story. Read up on Euclid, and his "Eureka" moment.

But, in short: Drop it in a container of water that is filled to the brim. Water will overflow, equal to the volume of the ring. Measure that.

Then, do it again, with a chunk of 24 karat gold that weighs the same. Measure the water that overflows again.

If the same amount of water overflows both times, the ring is 24 karats, because it must be the exact same weight and density. If the water amounts are different, then the ring has a different density, meaning the composition is different.

Make sense?
Brittany

Dalton, GA

#47 Sep 10, 2009
Iron rusts by reacting with oxygen in the presence of water according to the equation
4Fe(s)+ 3O2(g) 2Fe2O3(s)
An iron bar weighed 602 g. After the bar had been standing in moist air for a month, exactly one-eighth of the iron turned to rust. Calculate the final mass of the iron bar and rust.
Eva

Norristown, PA

#48 Sep 13, 2009
A gold colored nugget wass found that had a mass of 44.1 g. When placed in a graduate cylinder containing 32.0 mL of water, tha volume rose to 40.8 mL. Was the nugget gold ( density=19.3 g/mL) or fool's gold, iron pyrite (density=5.0 g/mL)?
Shellbee

Mount Juliet, TN

#49 Sep 22, 2009
In an experiment to determine how to make sulfur trioxide, a chemist combines 32.0 g of sulfur with 50.0g of oxygen. She finds that she made 80.0 of sulfur trioxide and had 2.0g left over oxygen. how would the chemist make 150.0g of sulfur trioxide so that she has no leftovers?
Eva

Norristown, PA

#50 Oct 4, 2009
How many electrons are in Na+ cation?
Pokay1kaduB

Bolivar, OH

#51 Oct 9, 2009
Eva wrote:
How many electrons are in Na+ cation?
One less than the sodium atom
Pokay1kaduB

Bolivar, OH

#52 Oct 9, 2009
Eva wrote:
A gold colored nugget wass found that had a mass of 44.1 g. When placed in a graduate cylinder containing 32.0 mL of water, tha volume rose to 40.8 mL. Was the nugget gold ( density=19.3 g/mL) or fool's gold, iron pyrite (density=5.0 g/mL)?
The information given provides you with the volume of the specimen which is
40.8ml - 32ml = 8.2ml (or 8.2cc, since 1cc = 1ml)

So the density is mass/volume. 44.1g/8.2ml

Can you do the rest? It is pyrite but can you tell me how to do the rest of the calculation to show it is indeed pyrite?
Pokay1kaduB

Bolivar, OH

#53 Oct 9, 2009
Brittany wrote:
Iron rusts by reacting with oxygen in the presence of water according to the equation
4Fe(s)+ 3O2(g) 2Fe2O3(s)
An iron bar weighed 602 g. After the bar had been standing in moist air for a month, exactly one-eighth of the iron turned to rust. Calculate the final mass of the iron bar and rust.
602/8 = 75.25 so 526.75 g is still iron.
75.25g has turned into iron oxide. SO you have to figure out how many moles of iron 75.25 grams is equivalent to.
Then from the equation you can see that two moles of iron are required to make one mole of iron oxide.(the formula for iron is Fe and the formula for iron oxide is Fe2O3 so we can also say the formula for iron is Fe1 so we can see that 1mol of iron oxide (Fe2)contains two moles of iron (Fe1 X 2 = Fe2)).

SO that means there is half the number of moles of iron oxide as there is iron. Once you figure out how many moles of iron 75.25 grams is, then you divide that number by 2 and that is the number of moles of iron oxide. Then you work backwards using the molecular weight and the number of moles of iron oxide to find the mass of it. Then of course you add that to 526.75g.
Pokay1kaduB

Bolivar, OH

#54 Oct 9, 2009
Shellbee wrote:
In an experiment to determine how to make sulfur trioxide, a chemist combines 32.0 g of sulfur with 50.0g of oxygen. She finds that she made 80.0 of sulfur trioxide and had 2.0g left over oxygen. how would the chemist make 150.0g of sulfur trioxide so that she has no leftovers?
I'm guessing you have to use more sulfur then. You have to convert everything to moles. And you have to look at an equation, 3O2 + 2S ---> 2SO3
The moles of O2 has to be 3/2 the moles of S.
SO you need 3 moles O2 and 2 moles S or 1.5 moles O2 and 1 mole S or.......
Mallory

United States

#55 Oct 22, 2009
Determine the number of grams of CO2 present in a 3.23 L container, under 2.87 atm, at 317.41 K.
Pokay1kaduB

Bolivar, OH

#56 Oct 31, 2009
Mallory wrote:
Determine the number of grams of CO2 present in a 3.23 L container, under 2.87 atm, at 317.41 K.
I always find these questions late but the answer is
found by first finding the moles of CO2 from the ideal gas law,
PV = nRT
I would supppose. The units are already correct for use in the equation, just substitute in the numbers. The question gives you P, T and V and you can look up the constant R (it's like 0.0821 and it's derived by solving the equation for R using 1 mol of an ideal gas at STP, but you don't have to derive that, it is given in your textbook I'm sure).
Then once you have the number of moles you simply convert it to grams using value for the molecular weight (44 grams/mol).
Ntwnangel

Hartford, CT

#57 Nov 9, 2009
Can someone help me with this problem?

Consider the square pyramidal AB5 molecule:
Using the correct character table, determine the possible hybridization schemes for the central atom A bonding to the symmetry adapted set of 5B ligand atoms based on the reducible representation obtained for the 5B atoms.
Gabby

Placentia, Canada

#58 Nov 16, 2009
calculate the volume of hydrogen gas that should be produced when 3.77g of magnesium completly reacts with excess hydrochloric acid at STP
conditions?
Pokay1kaduB

Bolivar, OH

#59 Nov 22, 2009
Ntwnangel wrote:
Can someone help me with this problem?
Consider the square pyramidal AB5 molecule:
Using the correct character table, determine the possible hybridization schemes for the central atom A bonding to the symmetry adapted set of 5B ligand atoms based on the reducible representation obtained for the 5B atoms.
http://www.springerlink.com/co ntent/x958547684863632/

Other than that I cannot help very much, I would have to dig into my books and spend some time reviewing to be able to help. That's one area I am rusty in.
Pokay1kaduB

Bolivar, OH

#60 Nov 22, 2009
Gabby wrote:
calculate the volume of hydrogen gas that should be produced when 3.77g of magnesium completly reacts with excess hydrochloric acid at STP
conditions?
2Mg + 2HCl --> H2 + 2MgCl
I just wrote the equation down to get a better picture but it is not necessary.
Convert 3.77g Mg to moles Mg
then use the fact that 1 mole of an ideal gas takes up 22.4 liters of volume
Pam

Alaska Peninsula Nwr, AK

#62 Dec 7, 2009
Can anybody help withthis? A 6.00-g quantity of a diprotic acid was dissolved in water and made up to exactly 225 mL. Calculate the molar mass of the acid if 25.0 mL of this solution required 10.5 mL of 1.00 M (molar) KOH for neutralization. Assume that both protons of the acid were titrated.

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