Answer:
Stretching is the lengthening and contraction is the shortening
Explanation:
Unless it's asking for eccentric as the stretching and concentric as the shortening
Answer:
a weighing balance, a measuring cylinder, a spatula, a beaker/flask, and a stirrer
Explanation:
The lab apparatus that would be needed to prepare a solution of sodium chloride would be <em>a weighing balance, a measuring cylinder, a spatula, a beaker/flask, and a stirrer.</em>
The weighing balance would be used to weigh out the required amount of sodium chloride. The beaker or flask would be placed on the weighing balance and its weight zeroed. The spatula would then be sued to take out the sodium chloride from its container into the beaker till the required amount is reached. The measuring cylinder would then be used to measure out the required volume of water which would be added to the salt in the beaker. The stirrer would then be used to stir the mixture in order for the salt to dissolve.
Given:
Diprotic weak acid H2A:
Ka1 = 3.2 x 10^-6
Ka2 = 6.1 x 10^-9.
Concentration = 0.0650 m
Balanced chemical equation:
H2A ===> 2H+ + A2-
0.0650 0 0
-x 2x x
------------------------------
0.065 - x 2x x
ka1 = 3.2 x 10^-6 = [2x]^2 * [x] / (0.065 - x)
solve for x and determine the concentration at equilibrium.
Answer: Solution A : ![[H_3O^+]=0.300\times 10^{-7}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D0.300%5Ctimes%2010%5E%7B-7%7DM)
Solution B : ![[OH^-]=0.107\times 10^{-5}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.107%5Ctimes%2010%5E%7B-5%7DM)
Solution C : ![[OH^-]=0.177\times 10^{-10}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.177%5Ctimes%2010%5E%7B-10%7DM)
Explanation:
pH or pOH is the measure of acidity or alkalinity of a solution.
pH is calculated by taking negative logarithm of hydrogen ion concentration and pOH is calculated by taking negative logarithm of hydroxide ion concentration.
![pH=-\log[H_3O^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH_3O%5E%2B%5D)
![[H_3O^+][OH^-]=10^{-14}](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%5BOH%5E-%5D%3D10%5E%7B-14%7D)
a. Solution A: ![[OH^-]=3.33\times 10^{-7}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D3.33%5Ctimes%2010%5E%7B-7%7DM)
![[H_3O^+]=\frac{10^{-14}}{3.33\times 10^{-7}}=0.300\times 10^{-7}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D%5Cfrac%7B10%5E%7B-14%7D%7D%7B3.33%5Ctimes%2010%5E%7B-7%7D%7D%3D0.300%5Ctimes%2010%5E%7B-7%7DM)
b. Solution B : ![[H_3O^+]=9.33\times 10^{-9}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D9.33%5Ctimes%2010%5E%7B-9%7DM)
![[OH^-]=\frac{10^{-14}}{9.33\times 10^{-9}}=0.107\times 10^{-5}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%5Cfrac%7B10%5E%7B-14%7D%7D%7B9.33%5Ctimes%2010%5E%7B-9%7D%7D%3D0.107%5Ctimes%2010%5E%7B-5%7DM)
c. Solution C : ![[H_3O^+]=5.65\times 10^{-4}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D5.65%5Ctimes%2010%5E%7B-4%7DM)
![[OH^-]=\frac{10^{-14}}{5.65\times 10^{-4}}=0.177\times 10^{-10}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%5Cfrac%7B10%5E%7B-14%7D%7D%7B5.65%5Ctimes%2010%5E%7B-4%7D%7D%3D0.177%5Ctimes%2010%5E%7B-10%7DM)
Answer:
Chemical change.
Explanation:
When it is heated it decomposes into mercury and oxygen gas. The mercury oxide reactant becomes the silver color of mercury. Hence, a color change can be noticed throughout the reaction.
