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nexus9112 [7]
3 years ago
14

¿Cuáles son las grasas e hidróxidos más efectivos en la elaboración de jabón?

Chemistry
1 answer:
d1i1m1o1n [39]3 years ago
6 0

Answer:

Las sales ideales para la saponificacion es la combinacion de grasas complejas con muchas cadenas de trigliceridos y de la soda caustia.

Explanation:

Un ejemplo de estos son la grasa vacuna o la aceite de cocina de consumo tradicional y la soda caustica que es el hidroxido de sodio.

Juntos estos dos productos forman el proceso de saponificacion.

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What could you predict about a methane molecule that would most likely be made by using both the quantum model of the atom and J
Svetlanka [38]
By using the quantum model of the atom and John Dalton's model you can predict that the methane molecule will most likely be made up of Hydrogen and Carbon.
5 0
3 years ago
A sample of an unknown liquid has a volume of 12.0 mL and a mass of 6 g. What is its density?
rjkz [21]

Answer:c

Explanation:

4 0
3 years ago
A gas has a volume of 25 ml, at a pressure of 1 atmosphere (1 atm). Increase the volume to 125 ml and the temperature remains co
vekshin1

<h2>0.2 atm </h2>

Explanation:

The new pressure can be found by using the formula for Boyle's law which is

P_1V_1 = P_2V_2 \\

Since we're finding the new pressure

P_2 =  \frac{P_1V_1}{V_2}  \\

We have

P_2  =  \frac{25 \times 1}{125}  =  \frac{25}{125}   = \frac{1}{5}  = 0.2 \\

We have the final answer as

<h3>0.2 atm</h3>

Hope this helps you

8 0
3 years ago
A proton is fired toward a lead nucleus from very far away. How much initial kinetic energy does the proton need to reach a turn
Olegator [25]

Answer:

The electric force is conservative.

Since

ΔK = −ΔU,

Kf − Ki =Ui −Uf.

We have,

Kf = 0

Ui = 0.

Thus Ki =Uf.

<u>so ,at 10 fm Uf = (2×10)−12 J.</u>

6 0
4 years ago
35.10 g of aluminum hydroxide is allowed to react with 53.94 g of sulfuric acid as follows:2Al(OH)3(s) + 3H2SO4(aq) --------&gt;
ikadub [295]

Answer:

62.586 gram

Explanation:

moles of Al(OH)3 = mass / molar mass = 35.1 / (27+17x3) = 0.45 mol

moles of H2SO4 = mass / molar mass = 53.94 / (2+32+16x4) = 0.55 mol

H2SO4 is the limiting reagent (reacts completely)

⇒ moles of Al2(SO4)3 is worked out by moles of H2SO4

moles of Al2(SO4)3 = moles of H2SO4 / 3 = 0.183 mol

mass of Al2(SO4)3 = mole x molar mass = 0.183 x (27x2 + 96x3) = 62.586 gram

8 0
3 years ago
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