All categories

4 years ago

When would it be most appropriate to use ppm to describe the concentration of the solutes in a solution?

Chemistry

1 answer:

cluponka [151]4 years ago

5 0

B) When describing very dilute solutions.

Explanation:

The most appropriate time to use ppm is when describing dilute concentrations of solutes in solutions. It is used mostly to determine trace amount of elements in water and soils.

The part per million ppm is the amount of miligram of a solute in a liter of solution.
It is similar to percentage.
Minute and trace concentrations are very difficult to work with.
Using parts per million, they become better appreciated and easy to use.

Learn more:

Parts per million brainly.com/question/2854033

#learnwithBrainly

You might be interested in

How many grams of oxygen are required to react with 12.0 grams of octane(C8H18) in the combustion of octane in gasoline?

solniwko [45]

Answer : 42 grams of oxygen are required.

Explanation :

Step 1 : Write balanced chemical equation.

The combustion reaction of octane with oxygen can be written as,

$2 C_{8}H{18} (g) + 25 O_{2} (g) \rightarrow 16 CO_{2} (g) +18 H_{2}O (g)$

Step 2 : Find moles of octane.

The given mass of octane is 12 g

Molar mass of octane is 114.2 g/mol

The moles of octane are calculated as,

$moles =\frac{grams}{MolarMass} =\frac{12g}{114.2g/mol} = 0.105 mol$

We have 0.105 mol octane.

Step 3: Use mole ratio from balanced equation to find moles of O₂

The mole ratio of octane and O₂ is 2 : 25. Let us use this as a conversion factor.

$0.105 mol C_{8}H_{18} \times\frac{25mol O_{2}}{2molC_{8}H_{18}} = 1.31 mol$

We have 1.31 mol O₂.

Step 4 : Convert moles of O₂ to grams.

The molar mass of O₂ is 32 g/mol.

The grams of O₂ can be calculated as,

$1.31 mol O_{2} \times\frac{32g}{mol} = 42 grams$

42 grams of oxygen are required.

8 0

3 years ago

What mass of aluminum (in g) would be required to completely react with 1.00 L of 0.450 M HBr in the following chemical reaction

Mama L [17]

Answer:

4.0473g. Attached in image are the calculations I used to reach this answer.

3 0

2 years ago

Aluminum flouride+ sulfuric acid=

Alexxx [7]

you get hydrofluoric acid

6 0

3 years ago

Help !! Im not sure how to do this

7nadin3 [17]

1)The third equation must be halved.
H20–>H2+1/2O2
So ΔΗ3=483.6/2
2)The second equation must be halved and reversed.
CO2–>CO+1/2O2
So ΔΗ2=-(-566)/2=566/2

And now :
C+O2–>CO2
CO2–>CO+1/2O2
H2O—>H2+1/2O2
(You add them)
C+H2O—>CO+H2
ΔΗrxn=ΔΗ1+ΔΗ2’+ΔΗ3’=-393.5+283+241,8=131,3

4 0

3 years ago

Appositives a force of 2 N to the left and 3 N to the right which is true of the Box motion

nexus9112 [7]

1N to the right s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s s ss s s s

3 0

3 years ago