Here we have to calculate the number of moles of valuable propane can be prepared from 1.8 moles of carbon.
From 1.8 moles of carbon 0.3 moles of propane can be prepared by the reaction.
From 6 moles of carbon (C) 1 moles of valuable propane (C₃H₈) can be prepared.
Thus from 1.8 moles of C we can obtain ×1.8 = 0.3 moles of the propane can be prepared.
Thus the amount of propane produced in this reaction is determined.
Answer:
Explanation:
molecular weight of N₂H₄ is as follows
2 x 14 + 4 x 1 = 32
so 32 gram of N₂H₄ is equal to one mole of N₂H₄
25 gram of N₂H₄ will be equal to 25 / 32 mole of N₂H₄
so numerical quantity needed to convert grams of N₂H₄ to mole is
1 / 32 or .03125 . This number needs to be multiplied with grams of N₂H₄
to convert it to number of moles.
Answer:
Total mass of the reactant = 2+2.5 =4.5 mg
Total mass of product = 4.15 mg
therefore, mass of unreacted oxygen = 4.50-4.15 = 0.35 g
Explanation:
1) Initial mass of the Cesium-137== 180 mg
Mass of Cesium after time t = N
Formula used :
Half life of the cesium-137 = = initial mass of isotope
N = mass of the parent isotope left after the time, (t)
= half life of the isotope
= rate constant
Now put all the given values in this formula, we get
Mass that remains after t years.
Therefore, the parent isotope remain after one half life will be, 100 grams.
2)
t = 70 years
N = 35.73 mg
35.73 mg of cesium-137 will remain after 70 years.
3)
N = 1 mg
t = ?
t = 224.80 years ≈ 225 years
After 225 years only 1 mg of cesium-137 will remain.