Pt195 having 78 atomic number is the daughter nucleus produced when Au195 undergoes electron capture.
An unstable isotopes can undergoes radioactive decay to form a more stable isotopes . Initial isotopes is called parent nuclide and the resultant isotopes from radioactive decay is called the daughter nuclide There are several radioactive decay like alpha decay , beta decay , gamma decay , electron capture etc .
In electron capture , the atom have only one less which is daughter isotopes than electron than atomic number of parent isotopes.
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Answer:
Fe(NO2)3
Explanation:
A poly atomic ion is an ion that contains more than one atom.
If we look at all the compounds that have been mentioned among the options; KBr and NaF are both ionic but do not contain any covalent poly atomic ion.
However, Fe(NO2)3 contains NO2^+ which is a poly atomic ion that contains the covalent bond.
Answer: Osmotic pressure : increases
Explanation:
Osmotic pressure is the minimum pressure which is applied to a solution to prevent the flow of solvent across a semipermeable membrane
= osmotic pressure
C= concentration in Molarity (number of moles of solute dissolved per liter of the solution)
R= solution constant
T= temperature
Thus as osmotic pressure is directly proportional top concentration, osmotic pressure will increase on increasing the concentration of a nonvolatile solute in water.
Answer: It is a Double Displacement(Replacement) Reaction.
Explanation:
Answer:
pH = 1.32
Explanation:
H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺
This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:
So first calculate the moles reacted and produced:
n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M
54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH
it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.
moles H₂M left = 0.074 - 0.015 = 0.059
moles HM⁻ produced = 0.015
Using the Henderson - Hasselbach equation to solve for pH:
ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325
Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.
For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.
