A solution must have both a solute and a solvent, and they must exist in solution together. the solute must be dissolved in the solvent; salt does not dissolve in a sugar bowl, nor does oil dissolve in water; it is a non polar molecule. we know that sugar (the solute) when added to water(solvent) will dissolve, creating a solution.
therefore, B
Answer: pH of resulting solution will be 13
Explanation:
pH is the measure of acidity or alkalinity of a solution.
Moles of 
ion = 
Moles of 
ion = 
For neutralization:
1 mole of ion will react with 1 mole of ion
0.01 mol of ion will react with =
of ion
Thus (0.012-0.01)= 0.002 moles of are left in 20 ml or 0.02 L of solution.
![[OH^-]=\frac{0.002}{0.02L}=0.1M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%5Cfrac%7B0.002%7D%7B0.02L%7D%3D0.1M)
![pOH=-log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-log%5BOH%5E-%5D)
![pOH=-log[0.1]=1](https://tex.z-dn.net/?f=pOH%3D-log%5B0.1%5D%3D1)
Thus the pH of resulting solution will be 13
Higher. Because this type of heat transfer is conduction, meaning that heat always transfers to cooler objects.
The ratio of mole number of the reactants and products is equal to the coefficients. So the answer is a. 18.75 mol. b. 35.1 g. c. 1.38 * 10^5 g.
Answer:
The answer is "2%"
Explanation:
Equation:
Formula:
![Ka = \frac{[H^{+}][NO_2^{-}]}{[HNO_2]}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5BNO_2%5E%7B-%7D%5D%7D%7B%5BHNO_2%5D%7D)
Let
![[H^{+}] = [NO_2^{-}] = x](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3D%20%5BNO_2%5E%7B-%7D%5D%20%3D%20x)
at equilibrium
therefore,
![[H^{+}] = 2.0\times 10^{-2} \ M = 0.02 \ M](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3D%202.0%5Ctimes%2010%5E%7B-2%7D%20%5C%20M%20%3D%200.02%20%5C%20M)
Calculating the % ionization:
![= \frac{([H^{+}]}{[HNO_2])} \times 100 \\\\= \frac{0.02}{1}\times 100 \\\\= 2\%\\\\](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B%28%5BH%5E%7B%2B%7D%5D%7D%7B%5BHNO_2%5D%29%7D%20%5Ctimes%20100%20%5C%5C%5C%5C%3D%20%5Cfrac%7B0.02%7D%7B1%7D%5Ctimes%20100%20%5C%5C%5C%5C%3D%202%5C%25%5C%5C%5C%5C)
