In an experiment, students were given an unknown mineral. The unknown mineral was placed in 150 ml of water. Once in the water,
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d. When aluminum-28 undergoes beta decay, silicon-28 is produced.
Explanation:
When the nuclei of aluminium-28 decays, it produces silicon- 28:
Aluminium ²⁸₁₃Al
Silicon 28 ²⁸₁₄Si
beta particle ⁰₋₁
²⁸₁₃Al → ²⁸₁₄Si + ⁰₋₁
This way, the mass and atomic number are conserved.
Conservation of mass number:
28 = 28 + 0, 28 = 28
13 = 14 -1 , 13 = 13
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Balancing nuclear equations brainly.com/question/10094982
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Explanation:
Formation of formate buffer will be as follows.
Formate buffer = HCOOH + HCOONa
Molar mass of HCOOH = 46 g/mol
Molar mass of HCOONa = 68 g/mol
of HCOOH = 3.75
As, = [HCOOH] + [HCOONa]
= [HCOONa] (here, dissociation of HCOOH is negligible)
Its preparation will be as follows.
750 ml of 0.25 M sodium formate buffer
Since, there are many combinations of salt + acid and among those possibilities one of them is as follows.
HCOONa = 0.25 M in 400 ml solution
HCOOH = x molat in 350 ml solution
Therefore as, pH =
4 = 3.75 +
x = 0.16 M
Therefore, molarity of the formic acid acid is 0.16 M.
Now, calculate the weight of sodium formate as follows.
= 6.8 g
And, number of milliliters of formic acid required is 350 ml.