Hello. Can I get someone to help me set up these problems?
1.) How many moles of N are there in 2.6 moles of NHO3?
2.) How many moles of C2H6 (g) are consumed when 4.4 mol CO2 (g) are produced?
3) How many milliliters of 0.80 M CuSO4 must be diluted to make 866 milliliters of 0.20 M CuSO4?
Also, can someone check these that I did:
4) What is the molarity of 969 mL aqueous solution containing 0.36 mol of Kl?
M=0.36 mol/.969L = 0.37
5) What is the empirical formula for a compound containing only carbon and hydrogen whose elemental analysis is 79.89% C and 20.11% H?
mol C = 6.65 mol
mol H = 19.95 mol
/6.65
= C1H3
4 and 5 look ok to me.
1) There is 1 mole of N in a mole of NHO3; therefore, 2.6 mols NHO3 will contain 2.6 mols N.
2)4.4 mols CO2 x (1 mol C/1 mol CO2)x(1 mol C2H6/2 mol C) = ??
3)M x mL = M x mL.
check my thinking.
Check my work.
Sure, I can help you set up these problems and check your work.
1) To find the number of moles of N in 2.6 moles of NHO3, you can use the mole ratio between N and NHO3. Since there is 1 mole of N in 1 mole of NHO3, you can simply say that there are 2.6 moles of N in 2.6 moles of NHO3.
2) To find the number of moles of C2H6 consumed when 4.4 moles of CO2 are produced, you can use the balanced chemical equation for the reaction. Assuming the balanced equation is:
2 C2H6 + 7 O2 → 4 CO2 + 6 H2O
You can see that for every 4 moles of CO2 produced, you need 2 moles of C2H6. So, you can set up a proportion:
4.4 moles CO2 * (2 moles C2H6 / 4 moles CO2) = ??
Simplifying the proportion, you get:
2.2 moles C2H6.
3) To find how many milliliters of 0.80 M CuSO4 should be diluted to make 866 milliliters of 0.20 M CuSO4, you can use the dilution equation:
M1V1 = M2V2
Where M1 = initial concentration, V1 = initial volume, M2 = final concentration, and V2 = final volume.
In this case:
M1 = 0.80 M, V1 = ? (unknown), M2 = 0.20 M, V2 = 866 mL.
Plugging these values into the equation, you get:
(0.80 M) * V1 = (0.20 M) * (866 mL)
Solving for V1:
V1 = (0.20 M * 866 mL) / 0.80 M
V1 ≈ 216 mL.
Regarding your work:
4) The calculation for the molarity of the aqueous solution containing 0.36 mol of Kl seems correct. The molarity is indeed 0.37 M.
5) The calculation for the empirical formula of the compound containing 79.89% C and 20.11% H looks fine as well. The empirical formula is C1H3.
Hope this helps! Let me know if you have any other questions.