How many moles of O2 should be supplied to burn 1.82 mol of C3H8 (propane) molecules In a camping stove?
1 answer:
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Answer : The half life of 28-Mg in hours is, 6.94
Explanation :
First we have to calculate the rate constant.
Expression for rate law for first order kinetics is given by:
where,
k = rate constant
t = time passed by the sample = 48.0 hr
a = initial amount of the reactant disintegrate = 53500
a - x = amount left after decay process disintegrate = 53500 - 10980 = 42520
Now put all the given values in above equation, we get
Now we have to calculate the half-life.
Therefore, the half life of 28-Mg in hours is, 6.94
Answer:
Explanation:
We want to convert from moles to grams, so we must use the molar mass.
<h3>1. Molar Mass</h3>The molar mass is the mass of 1 mole of a substance. It is the same as the atomic masses on the Periodic Table, but the units are grams per mole (g/mol) instead of atomic mass units (amu).
We are given the compound PI₃ or phosphorus triiodide. Look up the molar masses of the individual elements.
- Phosphorus (P): 30.973762 g/mol
- Iodine (I): 126.9045 g/mol
Note that there is a subscript of 3 after the I in the formula. This means there are 3 moles of iodine in 1 mole of the compound PI₃. We should multiply iodine's molar mass by 3, then add phosphorus's molar mass.
- I₃: 126.9045 * 3=380.7135 g/mol
- PI₃: 30.973762 + 380.7135 = 411.687262 g/mol
Use the molar mass as a ratio.
We want to convert 3.14 moles to grams, so we multiply by that value.
The units of moles of PI₃ cancel.
The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we calculated, that is the tens place.
- 1292.698
The 2 in the ones place tells us to leave the 9.
3.14 moles of phosphorous triiodide is approximately equal to <u>1290 grams of phosphorus triodide.</u>