Answer: The answer is 6.78 grams.
Explanation: The equation used for solving this type of problems is:
where, 
is the initial amount of radioactive substance, N is the remaining amount and n is the number of half lives.
Number of half lives is calculated on dividing the given time by the half life.
n = time/half life
Time is given as 48.0 hours and the half life is given as 4.536 days. let's make the units same and for this let's convert the half life from days to hours.
= 108.864 hours
So, 
= 0.441
Since 5.00 g is the required amount when the radioactive substance is delivered to the scientist, it would be the final amount that is N. We need to calculate the initial amount. Let's plug in the values in the equation:
= 6.78 g
So, 6.78 g of the radioactive substance needs to be ordered.
Answer:
220.42098 amu
Explanation:
(220 .9 X .7422) + (220 X .0.1278) + (218.1 X 0.13) = 220.42098 amu
These are weighted averages.
So, we will take mass of one and multiply by abundance percentage that is provided and add them together.
In order to calculate the average atomic mass, we have to convert the percentages of abundance to decimals. So, you get
(220 .9 X .7422) + (220 X .0.1278) + (218.1 X 0.13) = 220.42098 amu
Answer:
The molecular formula of the metal oxide is CuO
Explanation:
Molar mass 79.545 g/m belongs to CuO
This is the reaction:
2Cu(s) + O₂(g) → 2CuO (s)
2 mol of Cu produce 2 mol of CuO
Molar mass Cu = 63.5 g/m
Mass / Molar mass = 1.56 g / 63.5 g/m →0.0245 m
0.0245 mol of Cu produce 0.0245 mole of CuO
Molar mass CuO = 79.545 g/m
Mol . Molar mass = mass
0.0245 m . 79.545 g/m = 1.95 g
Answer:
The aromatic ringsare the number 1 and 2.
Identify which lone pairs are participating in resonance (in aromatic rings).
- One lone pair on the sulfur atom.
Explanation:
For a compound to be aromatic it has to comply with Hückel's Rule, which says that the quantity of electrons in the pi orbitals, has to be a multiple of
4n + 2
for n = 0, 1, 2, 3.
