<span>#1 is air radon, #2 is x-ray, #3 is ground, #4 is cosmic radiation, #5 is TV tube, #6 is weapons test fallout . That's all I got hope I helped!</span>
The number of Ml of C₅H₈ that can be made from 366 ml C₅H₁₂ is 314.7 ml of C₅H₈
<u><em>calculation</em></u>
step 1: write the equation for formation of C₅H₈
C₅H₁₂ → C₅H₈ + 2 H₂
Step 2: find the mass of C₅H₁₂
mass = density × volume
= 0.620 g/ml × 366 ml =226.92 g
Step 3: find moles Of C₅H₁₂
moles = mass÷ molar mass
from periodic table the molar mass of C₅H₁₂ = (12 x5) +( 1 x12) = 72 g/mol
moles = 226.92 g÷ 72 g/mol =3.152 moles
Step 4: use the mole ratio to determine the moles of C₅H₈
C₅H₁₂:C₅H₈ is 1:1 from equation above
Therefore the moles of C₅H₈ is also = 3.152 moles
Step 5: find the mass of C₅H₈
mass = moles x molar mass
from periodic table the molar mass of C₅H₈ = (12 x5) +( 1 x8) = 68 g/mol
= 3.152 moles x 68 g/mol = 214.34 g
Step 6: find Ml of C₅H₈
=mass / density
= 214.34 g/0.681 g/ml = 314.7 ml
Answer:
1. 12.6 moles
2. 8.95 moles
3. 2A + 5B → 3C
4. 48 moles
Explanation:
1. 2Fe + 3Cl₂ → 2FeCl₃
We assume the chlorine in excess. Ratio is 2:2
2 moles of Fe, can produce 2 moles of chloride
12.6 moles of Fe will produce 12.6 moles of chloride.
2. 2Fe + 3Cl₂ → 2FeCl₃
For the same reaction, first of all we need to convert the mass to moles:
500 g . 1mol / 55.85 g = 8.95 mol
As ratio is 2:2, the moles we have are the same, that the produced
4. The reaction for the combustion is:
2C₂H₆ (g) + 7O₂ (g) → 4CO₂ (g) + 6H₂O (l)
We assume the oxygen in excess.
Ratio is 2:6, so 2 mol of ethane produce 6 moles of water
Therefore 16 moles of ethane may produce (16 .6) / 2 = 48 moles
Answer:
I think it is 860 g MG Cl2
Explanation:
your welcome
<u>Given:</u>
Mass of Ag = 1.67 g
Mass of Cl = 2.21 g
Heat evolved = 1.96 kJ
<u>To determine:</u>
The enthalpy of formation of AgCl(s)
<u>Explanation:</u>
The reaction is:
2Ag(s) + Cl2(g) → 2AgCl(s)
Calculate the moles of Ag and Cl from the given masses
Atomic mass of Ag = 108 g/mol
# moles of Ag = 1.67/108 = 0.0155 moles
Atomic mass of Cl = 35 g/mol
# moles of Cl = 2.21/35 = 0.0631 moles
Since moles of Ag << moles of Cl, silver is the limiting reagent.
Based on reaction stoichiometry: # moles of AgCl formed = 0.0155 moles
Enthalpy of formation of AgCl = 1.96 kJ/0.0155 moles = 126.5 kJ/mol
Ans: Formation enthalpy = 126.5 kJ/mol