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bogdanovich [222]
3 years ago
9

A student uses a sample of KOH stock solution and dilutes it to a total of 120 mL. If the diluted solution is 0.60 M KOH and its

original concentration was 2.25 M, what was the volume of the original sample? *
1.4 mL

89 mL

32 mL

5.5 mL

(DON'T POST LINKS PLEASE)
Chemistry
1 answer:
Vesnalui [34]3 years ago
5 0

Answer:

5.5

Explanation:

i think so?????????

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A student has two compounds in two separate bottles but with no labels on either one. One is an unbranched alkane, octane, C8H18
Aleks [24]

Answer:

a) both substances are insoluble in water

b) both substances are soluble in ligroin

c) both substances suffer combustion, octane produces more CO₂ than hexene.

d) both substances are less dense than waterl, with hexene having the lowest density.

e) only hexene would react with bromine

f)  only hexene would react with permanganate

Explanation:

a) both substances are non-polar and water is polar

b) both substances are non-polar and lingroin is non-polar

c) C₈H₁₈ + 17.5O₂ → 8CO₂ + 9H₂O

    C₆H₁₂ + 9O₂ → 6CO₂ + 6H₂O

d) water = 997 kg/m³

    ocatne = 703 kg/m³

    hexene = 673 kg/m³

e) bromine test is used to detect unsaturations

f) permanganate test is used to detect unsaturations

7 0
3 years ago
 
Katyanochek1 [597]

Answer:

The answer to your question is     V2 = 4.97 l

Explanation:

Data

Volume 1 = V1 = 4.40 L                    Volume 2 =

Temperature 1 = T1 = 19°C               Temperature 2 = T2 = 37°C

Pressure 1 = P1 = 783 mmHg           Pressure 2 = 735 mmHg

Process

1.- Convert temperature to °K

T1 = 19 + 273 = 292°K

T2 = 37 + 273 = 310°K

2.- Use the combined gas law to solve this problem

                  P1V1/T1  = P2V2/T2

-Solve for V2

                  V2 = P1V1T2 / T1P2

-Substitution

                  V2 = (783 x 4.40 x 310) / (292 x 735)

-Simplification

                 V2 = 1068012 / 214620

-Result

                 V2 = 4.97 l

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