Answer:
8.3
Explanation:
pH is the measure of the H+ or H30 (they r the same thing) ions in a solution. it is equal to -log[H+]. [H+]= Molar concentration of H+ ions.
a. mass of iron = 69.92 g
b. percent yield = 93%
<h3>Further eplanation
</h3>
Percent yield is the compare of the amount of product obtained from a reaction with the amount you calculated
General formula:
Percent yield = (Actual yield / theoretical yield )x 100%
An actual yield is the amount of product actually produced by the reaction. A theoretical yield is the amount of product that you calculate from the reaction equation according to the product and reactant coefficients
a.
Reaction
Fe₂O₃+3CO⇒2Fe+3CO₂
MW Fe₂O₃ : 159.69 g/mol
mol Fe₂O₃
mol Fe₂O₃ : mol Fe = 1 : 2
mol Fe :
mass of Fe(Ar=55.845 g/mol) :
b.
actual yield = 65 g
theoretical yield = 69.92 g
percent yield :
Answer:
See explanation.
Explanation:
I highly suggest you watch OChem Tutor's videos on IUPAC nomenclature because the actual naming would take a lot of time to teach in text-based format. But here is how to name them:
1) I think there are two seperate pictures for number 1. The molecule on the left is 1-pentene and the one on the right is 4-methyl-1-pentene. If the whole thing is one molecule but there is just a bond missing where the red marker numbers are, that molecule would be 9-methyl-1,6-decadiene.
2) 4-methyl-2-pentene
3) 2,4-octadiene
4) 1,5-nonadiene
5) 2,5-dimethyl-3-hexene
6) 3,6-dimethyl-2,4-heptadiene
7) 2,5,5-trimethyl-2-hexene
The answer is 451.4124 argon
Answer:
0.486 L
Explanation:
Step 1: Write the balanced reaction
2 KCIO₃(s) ⇒ 2 KCI (s) + 3 O₂(g)
Step 2: Calculate the moles corresponding to 1.52 g of KCIO₃
The molar mass of KCIO₃ is 122.55 g/mol.
1.52 g × 1 mol/122.55 g = 0.0124 mol
Step 3: Calculate the moles of O₂ produced from 0.0124 moles of KCIO₃
The molar ratio of KCIO₃ to O₂ is 2:3. The moles of O₂ produced are 3/2 × 0.0124 mol = 0.0186 mol
Step 4: Calculate the volume corresponding to 0.0186 moles of O₂
0.0186 moles of O₂ are at 37 °C (310 K) and 0.974 atm. We can calculate the volume of oxygen using the ideal gas equation.
P × V = n × R × T
V = n × R × T/P
V = 0.0186 mol × (0.0821 atm.L/mol.K) × 310 K/0.974 atm = 0.486 L
