Explanation:
1) Initial mass of the Cesium-137=
= 180 mg
Mass of Cesium after time t = N
Formula used :
Half life of the cesium-137 = ![t_{1/2}=30 years[p/tex]where, [tex]N_o](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D30%20years%5Bp%2Ftex%5D%3C%2Fp%3E%3Cp%3Ewhere%2C%0A%3C%2Fp%3E%3Cp%3E%5Btex%5DN_o)
= initial mass of isotope
N = mass of the parent isotope left after the time, (t)
= half life of the isotope
= rate constant
Now put all the given values in this formula, we get
Mass that remains after t years.
Therefore, the parent isotope remain after one half life will be, 100 grams.
2)
t = 70 years
N = 35.73 mg
35.73 mg of cesium-137 will remain after 70 years.
3)
N = 1 mg
t = ?
t = 224.80 years ≈ 225 years
After 225 years only 1 mg of cesium-137 will remain.
Answer:
Answer below
Explanation:
First to show combustion, you use O₂.
So C₆H₁₀O₅ + O₂ is the reaction. Assuming complete combustion, all you get as products are CO₂ and H₂O. Then you have to balance the full reaction.
C₆H₁₀O₅ + 6O₂ -----> 6CO₂ + 5H₂O
6 Carbons on each side
10 Hydrogens on each side
17 Oxygens on each side
1.2 moles of (nph4)3po3 is.......159.6 grams
<u>Answer: </u>The chemical symbol of the element is Sulfur.
<u>Explanation:</u>
The element which is present in third period of the periodic table having four 3p electrons is Sulfur. Sulfur is the 16th element of the periodic table which has 6 valance electrons.
The electronic configuration of this element is: ![[Ne]3s^23p^4](https://tex.z-dn.net/?f=%5BNe%5D3s%5E23p%5E4)
This element is considered as a non-metal because it will accept electrons in order to attain stable electronic configuration.
Hence, the chemical symbol of the element is Sulfur.
