Answer : Both solutions contain
molecules.
Explanation : The number of molecules of 0.5 M of sucrose is equal to the number of molecules in 0.5 M of glucose. Both solutions contain
molecules.
Avogadro's Number is
=
which represents particles per mole and particles may be typically molecules, atoms, ions, electrons, etc.
Here, only molarity values are given; where molarity is a measurement of concentration in terms of moles of the solute per liter of solvent.
Since each substance has the same concentration, 0.5 M, each will have the same number of molecules present per liter of solution.
Addition of molar mass for individual substance is not needed. As if both are considered in 1 Liter they would have same moles which is 0.5.
We can calculate the number of molecules for each;
Number of molecules =
;
∴ Number of molecules =
which will be = 
Thus, these solutions compare to each other in that they have not only the same concentration, but they will have the same number of solvated sugar molecules. But the mass of glucose dissolved will be less than the mass of sucrose.
Mentos reacts to coca-cola because of the sugars and then they expand which causes the soda to explode out of the open container. this is a chemical reaction because it cause the sugars to expand which you can't see happen.
The reactant is what you begin with.
The product is what you end up with (so the answer is B)
Answer:
![[H^{+}] = 0.761 \frac{mol}{L}](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3D%200.761%20%5Cfrac%7Bmol%7D%7BL%7D)
![[OH^{-}]=1.33X10^{-14}\frac{mol}{L}](https://tex.z-dn.net/?f=%5BOH%5E%7B-%7D%5D%3D1.33X10%5E%7B-14%7D%5Cfrac%7Bmol%7D%7BL%7D)

Explanation:
HCl and HNO₃ both dissociate completely in water. A simple method is to determine the number of moles of proton from both these acids and dividing it by the total volume of solution.
. V_{HCl}(L) \\ n_{H^{+} } from HNO_{3} = [HNO_{3}](\frac{mol}{L}). V_{HNO_{3}}(L)](https://tex.z-dn.net/?f=n_%7BH%5E%7B%2B%7D%20%7D%20from%20HCl%20%3D%20%5BHCl%5D%28%5Cfrac%7Bmol%7D%7BL%7D%29.%20V_%7BHCl%7D%28L%29%20%20%5C%5C%20n_%7BH%5E%7B%2B%7D%20%7D%20from%20HNO_%7B3%7D%20%20%3D%20%5BHNO_%7B3%7D%5D%28%5Cfrac%7Bmol%7D%7BL%7D%29.%20V_%7BHNO_%7B3%7D%7D%28L%29)
Here, n is the number of moles and V is the volume. From the given data moles can be calculated as follows






For molar concentration of hydrogen ions:
![[H^{+}] = \frac{n_{H^{+}}(mol)}{V(L)}](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%20%3D%20%5Cfrac%7Bn_%7BH%5E%7B%2B%7D%7D%28mol%29%7D%7BV%28L%29%7D)
![[H^{+}] = \frac{0.761}{1.00}](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3D%20%5Cfrac%7B0.761%7D%7B1.00%7D)
![[H^{+}] = 0.761 \frac{mol}{L}](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3D%200.761%20%5Cfrac%7Bmol%7D%7BL%7D)
From dissociation of water (Kw = 1.01 X 10⁻¹⁴ at 25°C) [OH⁻] can be determined as follows
![K_{w} = [H^{+} ][OH^{-} ]](https://tex.z-dn.net/?f=K_%7Bw%7D%20%3D%20%5BH%5E%7B%2B%7D%20%5D%5BOH%5E%7B-%7D%20%5D)
![[OH^{-}]=\frac{Kw}{[H^{+}] }](https://tex.z-dn.net/?f=%5BOH%5E%7B-%7D%5D%3D%5Cfrac%7BKw%7D%7B%5BH%5E%7B%2B%7D%5D%20%7D)
![[OH^{-}]=\frac{1.01X10-^{-14}}{0.761 }](https://tex.z-dn.net/?f=%5BOH%5E%7B-%7D%5D%3D%5Cfrac%7B1.01X10-%5E%7B-14%7D%7D%7B0.761%20%7D)
![[OH^{-}]=1.33X10^{-14}\frac{mol}{L}](https://tex.z-dn.net/?f=%5BOH%5E%7B-%7D%5D%3D1.33X10%5E%7B-14%7D%5Cfrac%7Bmol%7D%7BL%7D)
The pH of the solution can be measured by the following formula:
![pH = -log[H^{+} ]](https://tex.z-dn.net/?f=pH%20%3D%20-log%5BH%5E%7B%2B%7D%20%5D)

