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kipiarov [429]
1 year ago
10

If 100 ml of 3. 0 m solution were diluted to 250 ml, what would the concentration be?.

Chemistry
1 answer:
AysviL [449]1 year ago
5 0

If o100 ml of 3. zero m solution were diluted to 250 ml awareness of<u> 1.2 M.</u>

The concentration of the solution tells you how tons solute has been dissolved within the solvent. for example, in case you upload one teaspoon to two cups of water, the awareness could be stated as 1 t salt in keeping with 2 c water.

V1 = one hundred mL, M1  = 3 M

V2 = 250 mL, M2 = ?

observe Dilution regulation

M1*V1 = M2*V2

so

M2 = M1*V1/V2

M2 = 100/250*3

M2 = 1.2 M

Concentration (normally uncountable, plural concentrations) The act, procedure, or capability of concentrating; the system of becoming focused, or the country of being concentrated. The course of interest to a particular object. The act, technique or product of reducing the quantity of a liquid,

Learn more about concentration here

brainly.com/question/17206790

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Definition empirical evidence
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Answer:

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3 years ago
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An aqueous KNO3 solution is made using 76.6 g of KNO3 diluted to a total solution volume of 1.84 L. (Assume a density of 1.05 g/
defon

Answer:

The answer to your question is Molarity = 0.41

Explanation:

Data

mass of KNO₃ = 76.6 g

volume = 1.84 l

density = 1.05 g/ml

Process

1.- Calculate the molecular mass of KNO₃

molecular mass = 39 + 14 + (16 x 3) = 101 g

2.- Calculate the number of moles

                      101 g of KNO₃  --------------- 1 mol

                       76.6 g of KNO₃ ------------  x

                        x = (76.6 x 1) / 101

                        x = 0.76 moles

3.- Calculate molarity

Molarity = \frac{number of moles}{volume}

Substitution

Molarity = \frac{0.76}{1.84}

Result

Molarity = 0.41

7 0
3 years ago
What ion must be present for a compound to be considered a base?
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It is OH- (hydroxide ion)
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On a recent trip to the Fort Worth Zoo, Trevor's parents bought him a helium-filled balloon. At the beginning of the day, the sk
timurjin [86]

Answer:

1. The new volume is 14L

2. The volume decreased

Explanation:

Step 1:

Data obtained from the question.

Initial Volume (V1) = 14.2 L

Initial pressure (P1) = 102.5 kPa

Initial temperature (T1) = 33°C

Final pressure (P2) = 100.9 kPa

Final temperature (T2) = 24°C.

Final volume (V2) =?

Step 2:

Conversion of celsius temperature to Kelvin temperature. This is illustrated below:

K = °C + 273

T1 = 33°C = 33°C + 273 = 306K

T2 = 24°C = 24°C + 273 = 297K

Step 3:

Determination of the new volume.

Applying the general gas equation P1V1/T1 = P2V2/T2, the new volume can be obtain as follow:

P1V1/T1 = P2V2/T2

102.5 x 14.2/306 = 100.9 x V2/297

Cross multiply to express in linear form as shown below:

306 x 100.9 x V2 = 102.5 x14.2 x297

Divide both side by 306 x 100.9

V2 = (102.5x14.2x297)/(306 x 100.9)

V2 = 14 L

The new volume is 14 L.

Step 4:

Determination of the change in volume. This is illustrated below

Final volume (V2) = 14 L

Initial volume (V1) = 14.2 L

Change in volume (ΔV) =?

Change in volume (ΔV) = Final volume (V2) - Initial volume (V1)

ΔV = V2 - V1

ΔV = 14 - 14.2

ΔV = - 0.2L

Since the change in volume is negative, it means there is a decrease in the volume.

7 0
3 years ago
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