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2 years ago

11

Which method would be best to separate a mixture of oil and water?

1 answer:

rosijanka [135]2 years ago

7 0

Answer:

distillation

Explanation:

Actually, distillation is used to separate liquids from nonvolatile solids, as in the separation of alcoholic liquors from fermented materials, or in the separation of two or more liquids having different boiling points, as in the separation of gasoline, kerosene, and lubricating oil from crude oil.

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Afirmar que el h20 es la formula que corresponde al agua, esto es una teoría, idea, ley, hipotesis o experimentación

frozen [14]

Answer:

Ley.

Explanation:

En la teoría de la ciencia, la regularidad de los procesos en la naturaleza se denomina ley de la naturaleza. Las leyes naturales se diferencian de otras leyes en que los seres humanos no pueden ponerlas en vigor ni anularlas a su discreción. En tal sentido, la composición química del agua es indudablemente una ley natural, en tanto el hombre no puede modificarla sin modificar las características inherentes del agua como tal.

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2 years ago

Which of the following describes a way in which chemistry affects how you

stepladder [879]

Answer:

I guess A, I am not sure...

6 0

2 years ago

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Can someone help please!

trasher [3.6K]

Answer:

Mixtures

Explanation:

7 0

2 years ago

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A 50.0 mL sample of a 1.00 M solution of CuSO4 is mixed with 50.0 mL of 2.00 M KOH in a calorimeter. The temperature of both sol

Reika [66]

Answer : The enthalpy change for the process is 52.5 kJ/mole.

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the solution

$q=[q_1+q_2]$

$q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]$

where,

q = heat released by the reaction

$q_1$ = heat absorbed by the calorimeter

$q_2$ = heat absorbed by the solution

$c_1$ = specific heat of calorimeter = $12.1J/^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_2$ = mass of water or solution = $Density\times Volume=1/mL\times 100.0mL=100.0g$

$\Delta T$ = change in temperature = $T_2-T_1=(26.3-20.2)^oC=6.1^oC$

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

$q=[(12.1J/^oC\times 6.1^oC)+(100.0g\times 4.18J/g^oC\times 6.1^oC)]$

$q=2623.61J$

Now we have to calculate the enthalpy change for the process.

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

where,

$\Delta H$ = enthalpy change = ?

q = heat released = 2626.61 J

n = number of moles of copper sulfate used = $Concentration\times Volume=1M\times 0.050L=0.050mole$

$\Delta H=\frac{2623.61J}{0.050mole}=52472.2J/mole=52.5kJ/mole$

Therefore, the enthalpy change for the process is 52.5 kJ/mole.

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3 years ago

In Europe, nutritional information is given in kilojoules (kJ) instead of nutritional calories (1 nutritional calorie = 1 kcal).

Alinara [238K]

There would be 68 calories per cup

8 0

2 years ago

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