The number of protons in an atom is equal the atomic number (= 50)
The sum of the Neutrons and Protons is the Atomic Mass (=125)
Neutrons + Protons=125
Plug in for the protons (50)
Neutrons +(50)=125
Then, once solved, we have Neutrons = 75
Assuming the atom is NOT an ion, the amount of electrons is equal to the number of protons. (protons=electrons=50=50)
Therefore:
There are 50 electrons, 50 protons, and 75 electrons.
Answer:
B. The only planet to support life.
Explanation:
Additional information
Relative atomic mass(Ar) : A=7, G=16
The empirical formula : A₂G
<h3>Further explanation</h3>
Given
3.5g of element A
4.0g of element G
Required
the empirical formula for this compound
Solution
The empirical formula is the smallest comparison of atoms of compound forming elements.
The empirical formula also shows the simplest mole ratio of the constituent elements of the compound
mol of element A :
mol of element G :
mol ratio A : G = 0.5 : 0.25 = 2 : 1
CxHy + O2 --> x CO2 + y/2 H2O
Find the moles of CO2 : 18.9g / 44 g/mol = .430 mol CO2 = .430 mol of C in compound
Find the moles of H2O: 5.79g / 18 g/mol = .322 mol H2O = .166 mol of H in compound
Find the mass of C and H in the compound:
.430mol x 12 = 5.16 g C
.166mol x 1g = .166g H
When you add these up they indicate a mass of 5.33 g for the compound, not 5.80g as you stated in the problem.
Therefore it is likely that either the mass of the CO2 or the mass of H20 produced is incorrect (most likely a typo).
In any event, to find the formula, you would take the moles of C and H and convert to a whole number ratio (this is usually done by dividing both of them by the smaller value).
The relationship of radiation with distance obeys the inverse square law. Therefore, doubling the distance decrease the radiation by a factor of 4. The new count is 250.
1) Applying the same principle, the count decreases by a factor of 100. The new count is 10
2) An alpha particle is 4He2 and the Hydrogen can be represented as 1H1
14N7 + 4He2 - 1H1
= 17X8
Proton number 8 belongs to Oxygen. Therefore, the resultant nucleus is:
17O8
3) 185Au79 - 4He2
= 181Ir77
4) X - 4He2 = 234Th90
X = 238U92
5) Beta emission results in the same nucleon number but an increase in the proton number; therefore, the result is:
234Pa91
