Answer:
The object A will be having the greater density compared to object B.
Explanation:
It is known that density of any object is defined as the mass of any object occupying a given volume. So the ratio of mass and volume will help to determine the density of any object. 
From the above equation, it can be seen that the density of any object is directly proportional to the mass of the object and inversely proportional to the volume occupied by the object.
So in the present context, the mass of objects A and B are same and it is 100 g. Thus, the density of object A and object B will be influenced by their volume. As it is given that the volume of object A is 50 cm3 and object B is 100 cm3, then depending upon the relationship of volume and density, the density of both the objects can be determined. As the object with higher volume will be having lesser density as volume is inversely proportional to density. Thus, in the given case the volume of object B is greater than object A and so the object A will be having greater density compared to object B.
Because of the common ion effect. Adding a common ion decreases the solubility of the salt which is the intention of a buffer.
Explanation:
⁶⁵₃₀Ca → ⁶⁵₂₉Sc + ⁿₓH
The reaction above is nuclear reaction.
In a nuclear reaction, the mass number and atomic number must be conserved.
The mass number is the superscript before the atom
Atomic number is the subscript before the atom
Conserving mass number:
65 = 65 + n
n = 0
conserving atomic number:
30 = 29 + x
x = 1
The unknown atom is a positron i.e a positively charged electron: ⁰₁e
⁶⁵₃₀Ca → ⁶⁵₂₉Sc + ⁰₁e
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FBI Fedral Burea of Investagation
Answer: -105 kJ
Explanation:-
The balanced chemical reaction is,

The expression for enthalpy change is,
![\Delta H=\sum [n\times B.E(reactant)]-\sum [n\times B.E(product)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Csum%20%5Bn%5Ctimes%20B.E%28reactant%29%5D-%5Csum%20%5Bn%5Ctimes%20B.E%28product%29%5D)
![\Delta H=[(n_{N_2}\times B.E_{N_2})+(n_{H_2}\times B.E_{H_2}) ]-[(n_{NH_3}\times B.E_{NH_3})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BN_2%7D%5Ctimes%20B.E_%7BN_2%7D%29%2B%28n_%7BH_2%7D%5Ctimes%20B.E_%7BH_2%7D%29%20%5D-%5B%28n_%7BNH_3%7D%5Ctimes%20B.E_%7BNH_3%7D%29%5D)
![\Delta H=[(n_{N_2}\times B.E_{N\equiv N})+(n_{H_2}\times B.E_{H-H}) ]-[(n_{NH_3}\times 3\times B.E_{N-H})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BN_2%7D%5Ctimes%20B.E_%7BN%5Cequiv%20N%7D%29%2B%28n_%7BH_2%7D%5Ctimes%20B.E_%7BH-H%7D%29%20%5D-%5B%28n_%7BNH_3%7D%5Ctimes%203%5Ctimes%20B.E_%7BN-H%7D%29%5D)
where,
n = number of moles
Now put all the given values in this expression, we get
![\Delta H=[(1\times 945)+(3\times 432)]-[(2\times 3\times 391)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%281%5Ctimes%20945%29%2B%283%5Ctimes%20432%29%5D-%5B%282%5Ctimes%203%5Ctimes%20391%29%5D)

Therefore, the enthalpy change for this reaction is, -105 kJ