i am diluting 16 grams of CaCl2 with 64 grams water. and ye i actually ment 3 instead of 2 (a typo) sorry. and i keep getting the same 1.85...
i tried this 5 times and i get the same answer. I checked my work. can you find another error. thank you  so much

I need help with this combustion problem i dont even know where to begin.

A 1.50 gram sample of cadaverine gives 3.23g CO2, 1.58g of N2O5,and 1.865g H2O. Its molar mass is 102.2 g/mol. Determine the empirical and molecular formulas.

Can some one please explain how to do the following problem:

How many grams would a 0.050 mol sample of silver (Ag) contain?

Thanks so much, it seems so simple now. I think I was just over analyzing it the whole time.

In general triple bonds are stronger than double bonds which in turn are stronger than single bonds.
A fair comparison would be C-C = 347  kJ/ mole, C=C =  612 kJ /mole; C=-C(triple bond)  =  838 kJ /mole. In these examples, a double bond is not quite twice as strong as a single bond, nor is a triple bond 3 times stronger than a single bond. This is explained by there being different types of bonds in the molecule (but beyond the scope of this level)

N2 which has a triple bond =  941 kJ /mole, O2  (O=O)  = 498 kJ /mole; H2 (H-H) = 436  kJ /mole

However, it is not a fair comparison here as the atoms are different in size, nuclear charge and internuclear distance.

Aluminium hydroxide is an ionic compound 

It came to me as a surprise that nitrogen has strongest bond. I learned that single bonds are strongest covalent bond. So why nitrogen having triple bonds has the strongest bond in diatomic molecules

Is Aluminium Hydroxide an alkali?

Thanks,

Ron

Is the CaCl2 a solution? sounds like. Or is it that you are adding enough solution that contains 16g?
16g of CaCl2 with density of 1.18g/ml

Assuming you are adding 16g of solution. This will have a volume of 13.56 mL  (V = MASS/DENSITY) and give a combined volume in water of 64 + 13.56 = 77.56mL
So I agree with you total volume but disagree with the volume of the added solution (0.01356L rather than 0.01256L)

Perhaps your error is in this step.

Find density of complete stopper (you will need the mass and volume for this). This will be the density of rubber and it will be the same for the drilled stopper (but volume will be less).

So if you know the volume of the drilled stopper use this and density to find the mass of the drilled stopper. (m = dv)

As you already have the mass of the complete stopper, subtract the mass of the drilled stopper to find the mass of the rubber removed.

I hope I have understood what you are asking