I've figured out number 7.
1) How would you prepare 30.0 ml of 0.600M HNO3 From a stock solution of 4.00M HNO3? What volume in ML of the 4.00 M HNo3 will you need?
c1v1 = c2v2
c = concentration
v = volume
0.600*30 mL = 4.00 x ?mL
Solve for ?mL.
2) if a 83.5 & was repoarted for a reaction that could have produced 12.1 g of product? what was the actual yield?
Is that 83.5%?
%yield = 83.5 = (actual yield/theoretical yield)*100 . You know the %yield (83.5%), and theoretical yield (12.1 g), substitute and solve for actual yield.
How many grams of product can be produced when starting with 10.0g of Iron (II) chloride and 5.0 g of Cl
You need to clarify this question. What reaction? What is Cl?
starting with 4.72moles of co, calculate the number of moles c02 produced if there is enough oxygen gas to react with all the co
You need to be careful with caps. Also you need to learn to write sentences. All sentences start with a capital letter and they end with periods. Note that CO, Co, and co have different meanings.
2CO + O2 ==> 2CO2
4.72 mols CO x (2 mols CO2/2 mols CO) = ?
at 1.00 atm and 273 k, 0.270 l of a gas weights 0.430g. calculate the molar mass of the gas.
Use PV = nRT and solve for n = mols gas. Then n = grams/molar mass. You know n and grams, solve for molar mass.
what is the K4C combustion formula.
I'm in the dark. You need to clarify.
Thanks Dr.Bob. For number three. Cl is Chlorine. And for number 6 K4C is potassium carbide. I wanted the combustion formula for that. I guess what it creates when it undergoes combustion.
3. Fe + Cl2 ==> FeCl2
This is a limiting reagent problem. You know that because you are given amounts for BOTH reactants.
mols Fe = grams/molar mass
mols Cl2 = grams/molar mass
Using the coefficients in the balanced equation, convert mols Fe to mols FeCl2.
Do the same for mols Cl2 to mols FeCl2
It is likely that these values will not agree with each other which means one of them is wrong. The correct value in limiting reagent (LR) problems is ALWAYS te smaller value and the reagent producing that value is the LR.
Using the smaller value, convert to grams. g = mols x molar mass
Sorry I didn't realize it was a compound. K4C isn't one of those off the shelf chemicals we see every day. Probably the following is the combustion.
K4C + O2 ==> CO2 + K2O
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