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kenny6666 [7]
3 years ago
9

How does the separation of a mixture containing salt, sand and naphthalene support the law of conservation of matter

Chemistry
1 answer:
Elis [28]3 years ago
3 0

Answer:

Mixtures are not unique to chemistry; we encounter them on a daily basis. The food and drinks ... The mixture that will be separated in this experiment contains three components: Naphthalene ... Dissolving the table salt with water to extract it from the sand by filtration, and. 3. ... support the Law of Conservation of Matter.

Explanation:

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stiks02 [169]

Answer:

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5 0
3 years ago
Read 2 more answers
The boiling point of 100 milliliters of water at sea level, vapor pressure 760 mm hg is 100oc. if, under different conditions, t
eimsori [14]
It has to be D. the time needed to boil the water decreased

7 0
3 years ago
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Which will provide the most direct measure of pH by measuring the hydrogen ion concentration of a solution?
pogonyaev
PH meter is dircet so the answer is C

5 0
3 years ago
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Given the following equation: 2C6H6 + 1502 - 12CO2 + 6H20
Margarita [4]

Answer:

a) 30 moles

Explanation:

                           2C6H6 + 1502 -------> 12CO2 + 6H20

from reaction      2 mol       15 mol

given                    4.0 mol     x mol

x = 4.0*15/2 = 30. mol

5 0
2 years ago
Calculate the pH after 10 mL of 1.0 M sodium hydroxide is added to 60 mL of 0.5M acetic acid.
baherus [9]

Explanation:

It is given that the total volume is (10 mL + 60 mL) = 70 mL.

Also, it is known that M_{1}V_{1} = M_{2}V_{2}

Where,    V_{1} = total volume

               V_{2} = initial volume

Therefore, new concentration of CH_{3}COOH = \frac{M_{2}V_{2}}{V_{1}}

                                        = \frac{60 \times 0.5}{70}

                                        = 0.43 M

New concentration of NaOH = \frac{M_{2}V_{2}}{V_{1}}

                                               = \frac{10 \times 1.0}{70}

                                               = 0.14 M

So, the given reaction will be as follows.

              CH_{3}COOH + OH^{-} \rightarrow CH_{3}COO^{-} + H_{2}O

Initial:             0.43          0.14                     0

Change:          -0.14        -0.14                    0.14

Equilibrium:    0.29          0                       0.14

As it is known that value of pK_{a} = 4.74

Therefore, according to Henderson-Hasselbalch equation calculate the pH as follows.

           pH = pK_{a} + log \frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]}

                 = 4.74 + log \frac{0.14}{0.29}

                 = 4.74 + (-0.316)

                 = 4.42

Therefore, we can conclude that the pH of given reaction is 4.42.

6 0
3 years ago
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