<span>There are factors that are important to be determined for the composition of ocean water. First factor is the temperatue, by this you will be able to know the rate of evaporation of the sea. Next is the salinity, through this you will be able to know salty the sea is which will help you identify the last factor- density. The density is the most common because when there is more salt in the sea, it is less dense.</span>
The scheme is shown below, the steps involved are as follow,
Step one: Reduction: The carbonyl group of given compound on reduction using
Wolf Kishner reagent converts the carbonyl group into -CH₂- group.
Step two: Epoxidation: The double bond present in starting compound when treated with
m-CPBA (<span>meta-Chloroperoxybenzoic acid) gives corrsponding epoxide.
Step three: Reduction: The epoxide is reduced to alcohol on treatment with
Lithium Aluminium Hydride (LiAlH</span>₄)<span> followed by hydrolysis.
Step four: Oxidation: The hydroxyl group (alcohol) is
oxidized to carbonyl (ketonic group) using oxidizing agent
Chromic acid (H</span>₂CrO₄).
Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane
We have that energy=specific heat * change in temperature * mass. Thus, we have the final temperature (22) minus the initial temperature (55) to equal -33 as our change in temperature. Our specific heat is in J/g*C, so we're good with that because g stands for grams and the aluminium is measured in grams. As there are 10 grams of aluminum, we have

as our final temperature
An exothermic reaction would release energy and would therefore lose heat itself, while an endothermic reaction would absorb energy and gain heat. Therefore, losing heat would be an exothermic reaction
Feel free to ask further questions!
Answer:
H+(aq) + OH-(aq) → H2O(l)
Explanation:
Step 1: Data given
nitrious acid = HNO3
sodium hydroxide = NaOH
Step 2: The unbalance equation
HNO3(aq) + NaOH(aq) →NaNO3(aq) + H2O(l)
The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side (Ba^2+ and Br-), look like this:
H+(aq) + NO3-(aq) + Na+(aq) + OH-(aq) →Na+(aq) +NO3(aq) + H2O(l)
H+(aq) + OH-(aq) → H2O(l)