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
All answered above.
Sure, I can help you set up these problems and also check the ones you've already done.
1.) How many moles of N are there in 2.6 moles of NHO3?
To figure out the number of moles of N in NHO3, you need to look at the chemical formula of NHO3. It shows that there is 1 N atom for every 1 NHO3 molecule. Therefore, the number of moles of N is the same as the number of moles of NHO3, which is 2.6 moles.
2.) How many moles of C2H6 (g) are consumed when 4.4 mol CO2 (g) are produced?
To determine the number of moles of C2H6 consumed, you need to use the balanced equation for the reaction involving C2H6 and CO2. Let's assume the balanced equation is:
C2H6 + xO2 -> yCO2 + zH2O
From the equation, we see that there are two C atoms in one C2H6 molecule and one C atom in one CO2 molecule. Therefore, the ratio of moles of C2H6 to moles of CO2 is 2:1. Thus, for every 1 mol of CO2 produced, 2 moles of C2H6 are consumed. So, if 4.4 mol of CO2 are produced, the corresponding moles of C2H6 consumed would be 2 * 4.4 = 8.8 mol.
3) How many milliliters of 0.80 M CuSO4 must be diluted to make 866 milliliters of 0.20 M CuSO4?
To calculate the volume of the 0.80 M CuSO4 solution needed, we can use the equation:
C1V1 = C2V2
Where C1 is the initial concentration, V1 is the initial volume, C2 is the desired concentration, and V2 is the desired volume.
Let's substitute the given values:
C1 = 0.80 M
V1 (unknown)
C2 = 0.20 M
V2 = 866 mL = 866/1000 = 0.866 L
0.80 M * V1 = 0.20 M * 0.866 L
Now, solve for V1:
V1 = (0.20 M * 0.866 L) / 0.80 M
V1 ≈ 0.216 L = 216 mL
So, you would need to dilute 216 milliliters of the 0.80 M CuSO4 solution to make 866 milliliters of 0.20 M CuSO4.
Now, let's check the problems you've already done:
4) What is the molarity of a 969 mL aqueous solution containing 0.36 mol of Kl?
Molarity (M) is defined as moles of solute divided by liters of solution. You correctly calculated:
M = 0.36 mol / 0.969 L ≈ 0.37 M
So, your answer of 0.37 M is correct.
5) What is the empirical formula for a compound containing only carbon and hydrogen, with an elemental analysis of 79.89% C and 20.11% H?
To determine the empirical formula, we need to find the simplest ratio of elements in the compound. The percentages given represent the mass percentages of each element.
To convert the mass percentages to moles, assume a 100 g sample. This would give you 79.89 g of carbon and 20.11 g of hydrogen.
Next, convert the grams to moles using the molar masses of carbon (12.01 g/mol) and hydrogen (1.01 g/mol):
moles of C = 79.89 g / 12.01 g/mol ≈ 6.65 mol
moles of H = 20.11 g / 1.01 g/mol ≈ 19.95 mol
Next, divide both moles by the smallest number of moles, which is 6.65 mol:
empirical formula: C6.65 H19.95
To simplify the formula, we can round off the decimal values to the nearest whole number:
empirical formula: C7 H20
So, your answer of C1H3 is incorrect. The correct empirical formula is C7H20.
I hope this helps! If you have any further questions, feel free to ask.