Answer:
In 1 mol of Pb₃(PO₄)₄ occupies 1001.48 grams
Explanation:
This compound is the lead (IV) phosphate.
Grams that occupy 1 mole, means the molar mass of the compound
Pb = 207.2 .3 = 621.6 g/m
P = 30.97 .4 = 123.88 g/m
O = (16 . 4) . 4 = 256 g/m
621.6 g/m + 123.88 g/m + 256 g/m = 1001.48 g/m
We are told that KOH is being used to completely neutral H₂SO₄ according to the following reaction:
KOH + H₂SO₄ → H₂O + KHSO₄
If KOH can completely neutralize H₂SO₄, then there must be an equal amount of moles of each as they are in a 1:1 ratio:
0.025 L x 0.150 mol/L = .00375 mol KOH
0.00375 mol KOH x 1 mole H₂SO₄/1 mole KOH = 0.00375 mol H₂SO₄
We are told we have 15 mL of H₂SO₄ initially, so now we can find the original concentration:
0.00375 mol / 0.015 L = 0.25 mol/L
The concentration of H₂SO₄ being neutralized is 0.25 M.
There are several differences between a physical and chemical change in matter or substances. A physical change in a substance doesn't change what the substance is. In a chemical change where there is a chemical reaction, a new substance is formed and energy is either given off or absorbed.
Answer: <span>Based on giveninformation, the compound is a
<u>Poly Unsaturated</u> fatty acid.
Explanation: Fatty acids are carboxylic acids with -COOH functional group and a chain of hydrocarbons.
Fatty Acids are classified as,
i) Saturated Fatty Acids: Those Fatty Acids in which there is no double bond between the carbon chain are called as saturated fatty acids.
Examples: Lauric Acid </span>CH₃(CH₂)₁₀COOH<span>
Myristic Acid </span>CH₃(CH₂)₁₂COOH<span>
Palmitic Acid </span>CH₃(CH₂)₁₄COOH
ii) Unsaturated Fatty Acids: Those Fatty Acids in which contain a single double bond (mono-unsaturated) or more than one double bond (poly-unsaturated) d between the carbon atoms in a chain are called as unsaturated fatty acids.
Examples: <span>
Linoleic acid</span>
<span>
Vaccenic acid </span><span>
Palmitoleic acid
The saturated fatty acid containing 26 carbon atoms is called as
Cerotic acid. While cerotic acid containing a double bond at position 3 and 9 has a IUPAC name
H</span>
exacosa-3,9-dienoic acid with following structure.
Answer:

Explanation:
Reactions:
With: ![K_1=\frac{[NO][O]}{[NO_2]}=6.8*10^{-49}](https://tex.z-dn.net/?f=K_1%3D%5Cfrac%7B%5BNO%5D%5BO%5D%7D%7B%5BNO_2%5D%7D%3D6.8%2A10%5E%7B-49%7D)

With: ![K_2=\frac{[NO_2}{[O_3][NO]}=5.8*10^{-34}](https://tex.z-dn.net/?f=K_2%3D%5Cfrac%7B%5BNO_2%7D%7B%5BO_3%5D%5BNO%5D%7D%3D5.8%2A10%5E%7B-34%7D)
To achive the reaction we combine those two reactions:
The equilibrium constant:
