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statuscvo [17]
3 years ago
11

Alcohol and water have different melting points. which method would you use to separate them?

Chemistry
1 answer:
yuradex [85]3 years ago
4 0
The process of separating alcohol from water can be done in several different ways. The most familiar method is through heating the blended liquid. Since alcohol has a lower boiling temperature than water, it will rapidly become steam. It can then be condensed into a separate container. You can also freeze the alcoholic mixture, which allows for partial removal of the nonalcoholic components; what remains will be more rich in alcohol. Use ordinary table salt to separate isopropyl alcohol from water. The result will be a condensed isopropyl alcohol, not a drinking alcohol.
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ivolga24 [154]
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5 0
2 years ago
Which set of compounds illustrates the law of multiple proportions? so2, so3 h2o, h2o2 n2o, no, no2 all of these?
Leto [7]
<span>All of these. All the gases that are mentioned in each set identify with the law of multiple proportions since the all the compounds have oxygen ions in them. Law of multiple proportions is defined as formation of compound with oxide ions after the reaction of an element with oxygen.</span>
7 0
3 years ago
a 20ml sample of hcl was titrated with the 0.0220 M Naoh. to reach the endpoint required 23.72 mL of the NaOh. Calculate the mol
Y_Kistochka [10]

Answer:

Molarity of HCl=0.026092M

Explanation:

The equation for the reaction is;

HCl + NaOH ⇒ NaCl + H2O

Using the formular, \frac{C_{A}V_{A}}{C_{B}V_{B} }=\frac{nA}{nB}    ..........equ1

whereC_{A} is the concentration of Acid,

          V_{A} is volume of acid

          C_{B} is concentration of the base

          V_{B} is volume of the base

          nA is the number of moles of Acid

          nB is number of moles of base

nA = 1,    nB=1 , V_{A}=20ml, C_{B}=0.022M, V_{B}=23.72mL

we will input these values into equation1 to solve for C_{A}

\frac{C_{A}*20}{0.022*23.72}=\frac{1}{1}

C_{A}*20=0.022*23.72

C_{A}=0.026092M

7 0
3 years ago
5.20763 to three significant figures​
Sergio039 [100]

Answer:

5.21

Explanation: You can only have 3 digits, so you would round to the hundredths place

5 0
2 years ago
In the coal-gasification process, carbon monoxide is converted to carbon dioxide via the following reaction: CO (g) + H2O (g) ⇌
Oksana_A [137]

Answer: Equilibrium constant is 0.70.

Explanation:

Initial moles of  CO = 0.35 mole

Volume of container = 1 L

Initial concentration of CO=\frac{moles}{volume}=\frac{0.35moles}{1L}=0.35M

Initial moles of  H_2O = 0.40 mole

Volume of container = 1 L

Initial concentration of H_2O=\frac{moles}{volume}=\frac{0.40moles}{1L}=0.40M

equilibrium concentration of CO=\frac{moles}{volume}=\frac{0.18moles}{1L}=0.18M [/tex]

The given balanced equilibrium reaction is,

                            CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)

Initial conc.            0.35 M       0.40M       0     0

At eqm. conc.    (0.35-x) M   (0.40-x) M   (x) M    (x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[CO_2]\times [H_2O]}{[CO]\times [H_2O]}

K_c=\frac{x\times x}{(0.40-x)(0.35-x)}

we are given : (0.35-x)= 0.18

x = 0.17

Now put all the given values in this expression, we get :

K_c=\frac{0.17\times 0.17}{(0.40-0.17)(0.35-0.17)}

K_c=0.70

Thus the value of the equilibrium constant is 0.70.

5 0
3 years ago
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