A) Salt, because salt can easily dissolve in water. Oil would not dissolve or evaporate in water (think of an oil spill - does that oil dissappear?). Aluminum foil would definitely not dissolve in water, so it is not soluble.
Answer:
Explanation:
The metabolic pathway by which energy can be obtained from a fatty acid is called <u>"beta-oxidation"</u>. In this route, acetyl-Coa is produced by removing <u>2 carbons</u> from the fatty acid for each acetyl-Coa produced. In other words, for each round, 1 acetyl Coa is produced and for each round 2 carbons are removed from the initial fatty acid. Therefore, the first step is to calculate the <u>number of rounds</u> that will take place for an <u>18-carbon fatty</u> acid using the following equation:
Where "n" is the <u>number of carbons</u>, in this case "18", so:
We also have to calculate the amount of Acetyl-Coa produced:
Now, we have to keep in mind that in each round in the beta-oxidation we will have the <u>production of 1 
and 1 
</u>. So, if we have 8 rounds we will have 8 and 8 .
Finally, for the total calculation of ATP. We have to remember the <u>yield for each compound</u>:
-) 
-) 
-) 
Now we can do the total calculation:
We have to <u>subtract</u> "2 ATP" molecules that correspond to the <u>activation</u> of the fatty acid, so:
In total, we will have 128 ATP.
I hope it helps!
Answer:
a) HNO3 -> H+ + NO3- disassociation of Nitric Acid; to yield a Nitrate ion and a Proton, H+, or as a Hydronium ion H3O+
b) H2S04 -> Disassociation of Sulfuric Acid; simple way- 2H+ + SO4- -
c) H2S hydrogen sulphide in water is an acid; thus H+ HS- disassociation.
d) NaOH -> dissociation of Na+ + OH-; this is complete; sodium hydroxide is deliquescent, meaning it will draw water - EVEN from the air! Strong Base
e) Na2CO3 -> 2Na+ CO3- - Ionization of sodium carbonate - a salt
f) Na2S04 -> 2Na+ + SO4 - - ionization of sodium sulphate - a salt
g) NaCl -> Na+ + Cl- ionization of the salt, Sodium Chloride
Explanation:
Salts ionize at different rates; acids or bases dissociate; these are mostly strong acids and NaOH, a strong base.
Answer:
62.98 % of the sample of hydrate is water
Explanation:
Step 1: Data given
Mass of the sample of a hydrate of sodium carbonate (Na2CO3) = 2.026 grams
After heating, the mass of the sample is 0.750 g
Molar mass H2O = 18.02 g/mol
Step 2: Calculate mass of water
Mass water = mass of hydrate - mass of sample after heating
Mass water = 2.026 grams - 0.750 grams
Mass water = 1.276 grams
Step 3: Calculate mass % percent of water
Mass % of water = (mass of water / total mass hydrate) * 100 %
Mass % of water = (1.276 grams / 2.026 grams) *100 %
Mass % of water = 62.98 %
62.98 % of the sample of hydrate is water
- Standard reduction potential of Ag/Ag⁺ is 0.80 v and that of Cu⁺²(aq)/Cu⁰ is +0.34 V.
- The couple with a greater value of standard reduction potential will oxidize the reduced form of the other couple.
Ag⁺ will be reduced to Ag(s) and Cu⁰ will be oxidized to Cu²⁺
Anode reaction: Cu⁰(s) → Cu²⁺ + 2 e⁻ E⁰ = +0.34 V
Cathode reaction: Ag⁺(aq) + e → Ag(s) E⁰ = +0.80 V
Cell reaction: Cu⁰(s) + 2 Ag⁺(aq) → Cu⁺²(aq) + 2 Ag⁰(s)
E⁰ cell = E⁰ cathode + E⁰ anode
= 0.80 + (-0.34) = + 0.46 V
