<h3>
Answer:</h3>
8 alpha particles
4 beta particles
<h3>
Explanation:</h3>
<u>We are given;</u>
- Neptunium-237
- Thallium-205
- Neptunium-237 undergoes beta and alpha decay to form Thallium-205.
We are required to determine the number of beta and alpha particles produced to complete the decay series.
- We need to know that when a radioisotope emits an alpha particle the mass number reduces by 4 while the atomic number decreases by 2.
- When a beta particle is emitted the mass number of the radioisotope increases by 1 while the atomic number remains the same.
In this case;
Neptunium-237 has an atomic number 93, while,
Thallium-205 has an atomic number 81.
Therefore;
²³⁷₉₃Np → x⁴₂He + y⁰₋₁e + ²⁰⁵₈₁Ti
We can get x and y
237 = 4x + y(0) + 205
237-205 = 4x
4x = 32
x = 8
On the other hand;
93 = 2x + (-y) + 81
but x = 8
93 = 16 -y + 81
y = 4
Therefore, the complete decay equation is;
²³⁷₉₃Np → 8⁴₂He + 4⁰₋₁e + ²⁰⁵₈₁Ti
Thus, Neptunium-237 emits 8 alpha particles and 4 beta particles to become Thallium-205.
Since the given solubility is 350 ppm, convert it first with fraction of solubility. by dividing the solubility with 10^6
S = 350 / 10^6
s = 3.5 x 10^-4
the multiply it to the total solution to calculate the amount of substance present
m = ( 3.5 x 10^-4 ) ( 1.01 )
m = 3.535 x 10^-4 g of the substance present
Answer:
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Explanation:
Answer:
Final T° is 45.5°C
Explanation:
Formula for calorimetry is:
Q = m . C . ΔT , where
ΔT = Final T° - Initial T°
C = Specific heat
m = mass
Let's replace with the data given
305 J = 93.4 g . 0.128 J/g°C . (Final T° - 20°C)
305 J / 93.4 g . 0.128 J/g°C = Final T° - 20°C
25.5°C = Final T° - 20°C → Final T° = 25.5°C + 20°C = 45.5°C
Answer:
empirical formula
This also means that in one mole of iso-octane, there are 8 moles of carbon and 18 moles of hydrogen. The empirical formula is the simplest whole-number ratio of atoms in the compound. For iso-octane, the empirical formula is C4H9. Notice that the molecular formula is twice that of the empirical formula.
