a few chemistry questions i really need help on...?

Question

so i have a final coming up in my chem. lab and i have been stumped on a few of the study material problems... by any chance can you help me with them? try to show your work as much as possible i need to figure out how to do them so i will be able to do them on the final as well. thanks everyone who helps me out with these...

1) A sample of 16.78 mL of 0.103 N H2SO4 is titrated with 0.177 N NaOH.

a) how many mL of NaOH will it take to reach the endpoint?
b) What is the molarity of the H2SO4 solution?
c)How many grams of NaOH will it take to reach the endpoint?

2) What volume will 4.4 grams of CO2 gas occupy over water at 30 derees Celcius and a barometric pressure of 700mm?

I will really appreciate any help offered with the two problems thank you once again!

Answer 1

Here is the approach

The acid base titration formula is M1V1 = M2V2 where M is molarity of the H in the acid and V is the volume.

You have data for N = normality. For NaOH it is the same as M. For H2SO4 1M = 2N, because the H2SO4 has 2 hydrogen ions as an acid.

So you can use the N values as is in the equation above, no adjustment required, and solve for the ml of NaOH. As mentioned, just divide the N by 2 to get the molarity of H2SO4

Once you have the ml of NaOH, divide by 1,000 and multiply by its molarity to get moles. Then multiply by 40 g/mol to get the grams

For problem 2, you need to use the ideal gas equation. COnvert the 4.4 grams to moles (its 0.1) and change the temp. to Kelvin. However, the pressure of the CO2 is not 700 mm. You have to subtract the partial pressure of water at 30 C. Look up that value. Once you have the real pressure of CO2, just plug in and solve for the volume.

Answer 2

1) Use this equation for all you titration & dilution problems:
V1 x C1 = V2 x C2 V1 is volume of compound 1 C1 conc. of compound 1 etc. N.B. if using M then need to factor in reactant ratios from equation e.g. as sulfuric has 2 hyrodrogens then 2 moles sod. hyd to fully neut. but with N then 1 :1

so a) 16.78 x 0.103 = V2 x 0.177 (V2 ml sod.hyd = 9.76ml)
b) [H2SO4] = 2 x .103 = 0.206 ( since have 2 replacable H's)
c) in a had 9.76 ml of 0.177 N NaOH this is = 9.76/1000 x 0.177 g =0.00173 g

2) use P x V = n x R x T (n.b. T ºK add 273 to ºC
R = 83.053 cm^3 atm K^-1 mol^-1
Mwt CO2=44

Answer 3

1a) 16.78 * 0.103 = 0.177 * V(NaOH)
V(NaOH) = 19.529 mL

b) H2SO4 --> 2 H+ + (SO4)2-
one mole of H2SO4 consists of 2 mole of H+, so:
molarity = 0.103 / 2 = 0.0515 M

c) NaOH --> Na+ + OH-
molarity = 0.177 / 1 = 0.177 M
mole = molarity * volume = 0.177 * 19.529 = 3.457 moles
mass = mole * molecular mass = 3.457 * 40 = 138.28 grams

2) n(CO2) = mole of CO2 = mass / molecular mass = 4.4 / 44 = 0.1 moles
p = pressure = 700 / 760 atm
T = temperature = (30 + 273) K = 303 K

pV = nRT
(700/760) * V = 0.1 * 0.08206 * 303
V = 2.7 L

Answer 4

a million.B (the only one that isn't an alkane.i.e. does not obey the formulation CnH2n+2) 2.B (all the others incorporate Oxygen- Hydrocarbons basically have Hydrogen and Carbon) 3.A (Has to obey formulation CnH2n-2 (alkynes)) 4.C (Pentene is an alkene....it incorporates one C=C bond) 5. it fairly is through fact it is an alkene - alkenes at the instant are not saturated as they incorporate double C=C bonds 6. 3 Pairs 3 bonds - Six electrons shared in finished