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3 years ago

Which unit can be used to express the concentration of a solution?

Chemistry

2 answers:

exis [7]3 years ago

6 0

Answer: (3) ppm

Explanation: Concentration is defined as the amount of solute dissolved in a known amount of the solvent or a solution.

Parts per million (ppm) is used to express the concentration when a very small quantity of solute is present in a large quantity of the solution. It is defined as the mass of solute present in one milion $(10^6)$ parts by mass of the solution.

$ppm=\frac{\text {mass of Solute}}{\text {Mass of solution}}\times 10^6$

L/s is used to express flow rate.

J/g is used to express energy per unit mass.

kPa is used to express pressure.

shutvik [7]3 years ago

5 0

Also to clarify further, L/s is a unit of flow rate, J/g is a unit of energy per unit mass, and kPa is a unit of pressure known as kilo pascals which is equivalent to 1000 pascals

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The reaction of C4H8 and Br2 to yield C4H2Br2 represents

natima [27]

Answer:

A Synthesis Reaction

Explanation:

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3 years ago

A gas with a volume of 250 mL at a temperature of 293K is heated to 324K. What is the new volume of the gas?

Natalija [7]

Answer:

The new volume of the gas is 276.45 mL.

Explanation:

Charles's law indicates that for a given sum of gas at constant pressure, as the temperature increases, the volume of the gas increases, and as the temperature decreases, the volume of the gas decreases.

Charles's law is a law that mathematically says that when the amount of gas and pressure are kept constant, the quotient that exists between the volume and the temperature will always have the same value:

$\frac{V}{T} =k$

Analyzing an initial state 1 and a final state 2, it is satisfied:

$\frac{V1}{T1} =\frac{V2}{T2}$

In this case:

V1= 250 mL
T1= 293 K
V2= ?
T2= 324 K

Replacing:

$\frac{250 mL}{293 K} =\frac{V2}{324 K}$

Solving:

$V2=324 K*\frac{250 mL}{293 K}$

V2= 276.45 mL

<em><u>The new volume of the gas is 276.45 mL.</u></em>

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3 years ago

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EastWind [94]

Answer

Rotating

Explanation:

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3 years ago

Read 2 more answers

A lightbulb filled with argon gas has a volume of 75 mL at STP. How many moles of argon gas does the lightbulb contain?

Elan Coil [88]

Answer:

0.00335 moles

Explanation:

From the question, Using

PV = nRT................... Equation 1

Where P = pressure, V = Volume, n = number of moles of argon gas, R = Molar gas constant, T = Temperature.

make n the subject of the equation

n = PV/RT............... Equation 2

Given: P = 1 atm (standard pressure), T = 273 K (standard temperature), V = 75 mL = 0.075 dm³

Constant: R = 0.082 atm·dm³/K·mol

Substitute into equation 2

n = (1×0.075)/(273×0.082)

n = 0.075/22.386

n = 0.00335 moles

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