The answer is in the picture which is given below:
Answer:
Last Quarter also called Third Quarter.
Explanation:
Answer:
The electronic configuration that are incorrectly written is 1s²2s³2p⁶, 4s²3d¹⁰4p⁷, 3s¹ and 2s²2p⁴.
Explanation:
The electronic configuration of the elements corresponds to how all the electrons of an element are arranged in energy levels and sub-levels.
There are 7 energy levels —from 1 to 7— whose sublevels are described as s, p, d and f.
All electronic configurations begin with the term "1s" —corresponding to the sublevel s of level 1— so 4s²3d¹⁰4p⁷, 3s¹ and 2s²2p⁴ are incorrectly written. In addition, 4s²3d¹⁰4p⁷ is written incorrectly because is impossible to jump from the sublevel "s" to the sublevel "d" —which is found from level 3 and up— without passing through the sublevel "p".
In the case of 1s²2s³2p⁶, the wrong thing is that the sublevel "s" can only hold two electrons, not three.
The other options are correctly written.
<h2>
Answer: 125.41 mL</h2>
Explanation:
Volume = mass ÷ density
= 116 g ÷ 0.925 g/mL
= 125.41 mL
<h3>A 116 g of sunflower oil of 0.925 g/mL has a volume of 125.41 mL.</h3>
Answer:
pH = 11.95≈12
Explanation:
Remember the reaction among aqueous acetic acid (
) and aqueous sodium hydroxide (NaOH)
First step. Need to know how much moles of the substances are present
![0.1 \frac{mol NaOH}{L} * 0.030 L = 0.003 mol NaOH[tex]0.1 \frac{mol CH_3COOH}{L} * 0.025 L= 0.0025 mol CH_3COOHSecond step. Know wich substance is in excess.0.0025 mol CH_3COOH * [tex]1 mol NaOH/ 1 mol CH_3COOH](https://tex.z-dn.net/?f=0.1%20%5Cfrac%7Bmol%20NaOH%7D%7BL%7D%20%2A%200.030%20L%20%3D%200.003%20mol%20NaOH%3C%2Fp%3E%3Cp%3E%5Btex%5D0.1%20%5Cfrac%7Bmol%20CH_3COOH%7D%7BL%7D%20%2A%200.025%20L%3D%200.0025%20mol%20CH_3COOH%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3ESecond%20step.%20Know%20wich%20substance%20is%20in%20excess.%3C%2Fp%3E%3Cp%3E0.0025%20mol%20CH_3COOH%20%2A%20%5Btex%5D1%20mol%20NaOH%2F%201%20mol%20CH_3COOH)
= 0.0025 mol NaOH
0.003 mol NaOH * 
/ 1 mol NaOH = 0.003 mol CH_3COOH[/tex]
NaOH is in excess. Now, how much?
0.003 mole NaOH - 0.0025 mole NaOH = 0.0005 mole NaOH
Then, that amount in excess would be responsable for the pH.
Third step. Know the pH
Remember that pH= -log[H+]
According to the dissociation of water equilibrium
Kw=[H+]*[OH-]= 10^(-14)
The dissociation of NaOH is
NaOH -> 
Now, concentration of OH^{-}[/tex] would be given for the excess of NaOH.
[OH-]= 0.0005 mole / 0.055 L = 0.00909 M
Careful: we have to use the total volumen
Les us to calculate pH
![pH= -log [H+]\\pH= -log \frac{K_w}{[OH-]} \\pH= 11.95](https://tex.z-dn.net/?f=pH%3D%20-log%20%5BH%2B%5D%5C%5CpH%3D%20-log%20%5Cfrac%7BK_w%7D%7B%5BOH-%5D%7D%20%5C%5CpH%3D%2011.95)
