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Dmitrij [34]
3 years ago
9

What term is used to describe alkaline hydrolysis of fats and oils​

Chemistry
1 answer:
Gnesinka [82]3 years ago
8 0

Answer:

Saponification

Explanation:

When treated with a base such as NaOH, the ester is transformed back to alcohol and the sodium salt of carboxylic acid. Saponification is the name <u>given</u> to the reaction since it is utilized in the production of soap.

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using the law of conservation of energy explain three types of energy transformations that occur during a roller coaster ride fr
AnnZ [28]

Answer:

energy a b c

Explanation:

5 0
3 years ago
Read 2 more answers
If a cube has a mass of 90.91 kilograms and a weight of 200 pounds on Earth, what will its mass and weight be on another planet?
Fudgin [204]

Answer:

The mass will be 90.91 kilograms, we can’t figure out the weight without knowing how much gravity is on the other planet.

Explanation:

Mass is the amount of substance a matter contains. The more the substance a matter contains, the more massive it becomes. The mass of an object is the same everywhere in the universe. Therefore, the cube will have the same mass of 90.91Kg in another planet.

Weight is a function of mass and the force of gravity on such a body. The weight of a body relies on the prevailing acceleration due to gravity in a particular place. Some places have gravity higher than that on earth and so they will have more weight. This is why we can't figure out weight without knowing the gravity in the other planet.

5 0
3 years ago
If 5 mol of oxygen gas effuses through an opening in 10 seconds, how long will it take for the same amount of hydrogen gas to ef
andrezito [222]

Answer:

B

Explanation:

Recall the law of effusion:

\displaystyle \frac{r_1}{r_2} = \sqrt{ \frac{\mathcal{M}_2}{\mathcal{M}_1} }

Because 5 mol of oxygen was effused in 10 seconds, the rate is 0.5 mol/s.

Let the rate of oxygen be <em>r</em>₁ and the rate of hydrogen be <em>r</em>₂.

The molecular weight of oxygen gas is 32.00 g/mol and the molecular weight of hydrogen gas is 2.02 g/mol.

Substitute and solve for <em>r</em>₂:

\displaystyle \begin{aligned} \frac{(0.5\text{ mol/s})}{r_2} & = \sqrt{\frac{(2.02\text{ g/mol})}{(32.00\text{ g/mol})}} \\ \\  r_2 & = \frac{0.5\text{ mol/s}}{\sqrt{\dfrac{(2.02\text{ g/mol})}{(32.00\text{ g/mol})}}} \\ \\ & = 2.0\text{ mol/s}\end{aligned}

Because there are 5 moles of hydrogen gas:

\displaystyle 5.0\text{ mol} \cdot \frac{1\text{ s}}{2.0\text{ mol}} = 2.5\text{ s}

In conclusion, it will take about 2.5 seconds for the hydrogen gas to effuse.

Check: Because hydrogen gas is lighter than oxygen gas, we expect that hydrogen gas will effuse quicker than oxygen gas.

6 0
2 years ago
When a 2.00 g sample of KCl is dissolved in water in a calorimeter that has a total heat capacity of 1.28 kJ ⋅ K − 1 , the tempe
Alchen [17]

Answer : The molar heat of solution of KCl is, 17.19 kJ/mol

Explanation :

First we have to calculate the heat of solution.

q=c\times (\Delta T)

where,

q = heat produced = ?

c = specific heat capacity of water = 1.28kJ/K

\Delta T = change in temperature = 0.360 K

Now put all the given values in the above formula, we get:

q=1.28kJ/K\times 0.360K

q=0.4608kJ=460.8J

Now we have to calculate the molar heat solution of KCl.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy change = ?

q = heat released = 460.8 J

m = mass of KCl = 2.00 g

Molar mass of KCl = 74.55 g/mol

\text{Moles of }KCl=\frac{\text{Mass of }KCl}{\text{Molar mass of }KCl}=\frac{2.00g}{74.55g/mole}=0.0268mole

Now put all the given values in the above formula, we get:

\Delta H=\frac{460.8J}{0.0268mole}

\Delta H=17194.029J/mol=17.19kJ/mol

Therefore, the molar heat of solution of KCl is, 17.19 kJ/mol

7 0
3 years ago
If mercury (Hg) and oxygen (O2) were reacted to form mercury oxide how many molecules of each reactant and product would be need
dimulka [17.4K]
In order to get HgO you would need 2Hg+1O2=2HgO. Since oxygen is diatomic you need two when it stands alone causing you to need two mercuries to balance out the reactants and the product I hope this helps
5 0
3 years ago
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