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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 is mineral in science?​
Klio2033 [76]

Answer:

minerals

Explanation:

defines a mineral as "a naturally occurring inorganic element or compound having an. orderly internal structure and characteristic chemical composition, crystal form, and physical. properties." Minerals differ from rocks, which are naturally occurring solids composed of one or more minerals.

6 0
3 years ago
2.Which term best describes the processes that change rocks and mountains on Earth’s surface over time?
ArbitrLikvidat [17]

Answer:

B. weathering

or erosion

4 0
2 years ago
Assume that the variables x and y are inversely related. If k = 18, what is the value of y for each of the following points?
kogti [31]
6_____18 record your data to be used in the following problem.
4 0
3 years ago
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The picture shows a natural environment.
Marta_Voda [28]

Answer: The answer is B

Explanation: cause the bush and cactus are plants and they are living things and rat and snake are animal which is also living thing

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3 years ago
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The balanced combustion reaction for C6H6 is 2C6H6(l)+15O2(g)⟶12CO2(g)+6H2O(l)+6542 kJ If 6.200 g C6H6 is burned and the heat pr
aliina [53]

Answer:

31.9178 °C is the final temperature of the water

Explanation:

2C_6H_6(l)+15O_2(g)\rightarrow 12CO_2(g)+6H_2O(l)+6542 kJ

Mass of benzene burned = 6.200 g

Moles of benzene burned = \frac{6.200 g}{78 g/mol}=0.0794 mol

According to reaction , 2 moles of benzene gives 6542 kJ of energy on combustion.

Then 0.0794 mol of benzene on combustion will give:

\frac{6542 kJ}{2}\times 0.0794 kJ=259.7174 kJ=Q

Mass of water in which Q heat is added = m = 5691 g

Initial temperature = T_i=21^oC

Final temperature = T_f

Specific heat of water = c = 4.18 J/g°C

Change in temperature of water = T_f-T_i

Q=mc\Delta t=mc(T_f-T_i)

259,717.4 J=5691 g\times 4.18 J/g^oC\times (T_f-21^oC)

T_f=31.91 ^oC

31.9178 °C is the final temperature of the water

5 0
3 years ago
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