Answer:
I'm pretty sure it's only 5, because once you've used up all 5 marshmallows, you can't make any more s'mores. Unless just graham crackers and chocolate counts as a s'more...
:)
It’s 5.88 x 10^-23.............
Answer:
(1) Magnesium carbonate :-
![Q_{MgCO_3}=[Mg^{2+}][CO_3^{2-}]](https://tex.z-dn.net/?f=Q_%7BMgCO_3%7D%3D%5BMg%5E%7B2%2B%7D%5D%5BCO_3%5E%7B2-%7D%5D)
(2) Calcium phosphate :-
![Q_{Ca_3(PO_4)_2}=[Ca^{2+}^3][PO_4^{3-}^2]](https://tex.z-dn.net/?f=Q_%7BCa_3%28PO_4%29_2%7D%3D%5BCa%5E%7B2%2B%7D%5E3%5D%5BPO_4%5E%7B3-%7D%5E2%5D)
(3) Silver sulfide :-
![Q_{Ag_2S}=[Ag^{+}^2][S^{2-}]](https://tex.z-dn.net/?f=Q_%7BAg_2S%7D%3D%5BAg%5E%7B%2B%7D%5E2%5D%5BS%5E%7B2-%7D%5D)
Explanation:
(1) Magnesium carbonate will dissociate as:-

The expression for the ion-product is as follows:-
![Q_{MgCO_3}=[Mg^{2+}][CO_3^{2-}]](https://tex.z-dn.net/?f=Q_%7BMgCO_3%7D%3D%5BMg%5E%7B2%2B%7D%5D%5BCO_3%5E%7B2-%7D%5D)
(2) Calcium phosphate will dissociate as:-

The expression for the ion-product is as follows:-
![Q_{Ca_3(PO_4)_2}=[Ca^{2+}^3][PO_4^{3-}^2]](https://tex.z-dn.net/?f=Q_%7BCa_3%28PO_4%29_2%7D%3D%5BCa%5E%7B2%2B%7D%5E3%5D%5BPO_4%5E%7B3-%7D%5E2%5D)
(3) Silver sulfide will dissociate as:-

The expression for the ion-product is as follows:-
![Q_{Ag_2S}=[Ag^{+}^2][S^{2-}]](https://tex.z-dn.net/?f=Q_%7BAg_2S%7D%3D%5BAg%5E%7B%2B%7D%5E2%5D%5BS%5E%7B2-%7D%5D)
If you are asking which is the most abundant, Uranium-238 is
Answer: Option D is correct.
Explanation: Equation given by de Broglie is:

where,
= wavelength of the particle
h = Planck's constant
m = mass of the particle
v = velocity of the particle
In option A, football will have some mass and is moving with a velocity of 25 m/s, hence it will have some wavelength.
In Option B, unladen swallow also have some mass and is moving with a velocity of 38 km/hr, hence it will also have some wavelength.
In Option C, a person has some mass and is running with a velocity of 7 m/hr, hence it will also have some wavelength.
As, the momentum of these particles are large, therefore the wavelength will be of small magnitude and hence, is not observable.
From the above, it is clearly visible that all the options are having some wavelength, so option D is correct.