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.
<span>Here we are asked to know the type of bond
between a glycosidic bond. A glycosidic bond is a type of bond that
exists between a carbohydrate molecule to another carbohydrate molecule. A
glycosidic bond creates between two monosaccharides can also be called as an
ether bond.</span>
Answer: 0.07868 mol H₂O
Explanation:
1) Chemical equation:
Cu₂O +H₂ → 2Cu + H₂O
2) mole ratios:
1 mol Cu₂O : 1 mol H₂ : 2 mol Cu : 1 mol H₂O
3) Convert 10.00 g of Cu to grams, using the atomic mass:
Atomic mass of Cu: 63.546 g/mol
number of moles = mass in grams / atomic mass = 10.00g / 63.546 g/mol
number of moles = 0.1574 mol
4) Use proportions
2mol Cu 0.1574 mol Cu
--------------- = ---------------------
1 mol H₂O x
⇒ x = 0.1574 mol Cu × 1 mol H₂O / 2mol Cu = 0.07868 mol H₂O
That is the answer
