Answer:
B) All carbon atoms have 6 neutrons.
Explanation:
The false statement from the given choices is that all carbon atoms have 6 neutrons.
There is a phenomenon called isotopy in chemistry.
Isotopy is the existence of two or more atoms of the same element having the same atomic number but different mass numbers due to the differences in the number of neutrons in their various nuclei.
- These atoms of elements are called isotopes.
- Carbon atoms generally have proton number of 6 which is the same as the atomic number.
- As with all atoms, the mass number or atomic mass equals the number of protons and neutrons.
For the isotopes of carbon, their number of neutrons differs.
For example:
<em>¹²₆C ¹³₆C ¹⁴₆C</em>
The number of neutrons differs in the above isotopes.
Number of neutrons = mass number - atomic number;
¹²₆C , number of neutrons = 12 - 6 = 6
¹³₆C, number of neutrons = 13 - 6 = 7
¹⁴₆C, number of neutrons = 14 - 6 = 8
Therefore, based on the concept of isotopy, all carbon atoms do not have 6 neutrons.
Answer:
8.625 grams of a 150 g sample of Thorium-234 would be left after 120.5 days
Explanation:
The nuclear half life represents the time taken for the initial amount of sample to reduce into half of its mass.
We have given that the half life of thorium-234 is 24.1 days. Then it takes 24.1 days for a Thorium-234 sample to reduced to half of its initial amount.
Initial amount of Thorium-234 available as per the question is 150 grams
So now we start with 150 grams of Thorium-234





So after 120.5 days the amount of sample that remains is 8.625g
In simpler way , we can use the below formula to find the sample left

Where
is the initial sample amount
n = the number of half-lives that pass in a given period of time.
From the electric generator to electric outlets in homes
Percentage yield = (actual yield / theoretical yield) x 100%
The balanced equation for the decomposition is,
2Na₃(CO₃)(HCO₃)·2H₂O(s) → 3Na₂CO₃(s) + CO₂(g) + 5H₂<span>
O(g)The
stoichiometric ratio between </span>Na₃(CO₃)(HCO₃)·2H₂O(s) and Na₂CO₃(s) is
2 : 3The decomposed mass of Na₃(CO₃)(HCO₃)·2H₂O(s) = 1000 kg
= 1000 x 10³ g
Molar mass of Na₃(CO₃)(HCO₃)·2H₂O(s) = 226 g mol⁻¹
moles of Na₃(CO₃)(HCO₃)·2H₂O(s) = mass / molar mass
= 1000 x 10³ g / 226 g mol⁻¹
= 4424.78 mol
Hence, moles of Na₂CO₃ formed = 4424.78 mol x

= 6637.17 mol
Molar mass of Na₂CO₃ = 106 g mol⁻¹
Hence, mass of Na₂CO₃ = 6637.17 mol x 106 g mol⁻¹
= 703540.02 g
= 703.540 kg
Hence, the theoretical yield of Na₂CO₃ = 703.540 kg
Actual yield of Na₂CO₃ = 650 kg
Percentage yield = (650 kg / 703.540 kg) x 100%
=
92.34%