M(NH₃)=51g, n(NH₃)=m/M=51g÷17g/mol(14+3×1)=3mol
N(NH₃)=n×Na(Avogadro number)=3mol×6·10²³1/mol=18×10²³ molecules
Answer:
in an oxygen atom there are:
protons:8
electrons:8
neutrons:8
Explanation:
this is because the atomic number of oxygen is 8 and that is the proton number and the electron number is the same as the atomic number
Answer:
For your first question, Curium does not occur naturally on Earth, meaning that it is not produced naturally on Earth. However, it can be formed in nuclear reactors.
For your second question, Curium has been used to provide power to electrical equipment used on space missions, but doesn't seem to be that important overall.
Explanation:
Hope this helped!
<span>a) 7.9x10^9
b) 1.5x10^9
c) 3.9x10^4
To determine what percentage of an isotope remains after a given length of time, you can use the formula
p = 2^(-x)
where
p = percentage remaining
x = number of half lives expired.
The number of half lives expired is simply
x = t/h
where
x = number of half lives expired
t = time spent
h = length of half life.
So the overall formula becomes
p = 2^(-t/h)
And since we're starting with 1.1x10^10 atoms, we can simply multiply that by the percentage. So, the answers rounding to 2 significant figures are:
a) 1.1x10^10 * 2^(-5/10.5) = 1.1x10^10 * 0.718873349 = 7.9x10^9
b) 1.1x10^10 * 2^(-30/10.5) = 1.1x10^10 * 0.138011189 = 1.5x10^9
c) 1.1x10^10 * 2^(-190/10.5) = 1.1x10^10 * 3.57101x10^-6 = 3.9x10^4</span>
Answer:
Atomic number of this isotope = 77
Explanation:
Given that,
Mass number = 193
No of neutrons = 116
We need to find the atomic no of this isotope.
We know that,
Atomic mass = No of protons + No. of neutrons
Also, atomic no = no of protons
So,
Atomic mass = atomic no + No. of neutrons
⇒ Atomic no = Atomic mass - no of neutrons
Atomic no = 193 - 116
Atomic no = 77
Hence, 77 is the atomic no of the isotope.
