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3 years ago

How much water would I need to add to 500 mL of a 2.4 M KCl solution to make a 1.0 M solution?

Chemistry

1 answer:

GalinKa [24]3 years ago

3 0

Answer:

700 mL of water

Explanation:

This is the perfect example of dilation calculations. Along with this concept we have a formula c = n ( solute ) / V ( solution ). Let us first solve for n by changing this equation to isolate the solute,

n ( KCL ) = 2.4 mol / L * 500 * 10 ^ - 3 L,

n ( KCL ) = 1.2 moles ( KCL )

Knowing the amount of moles of potassium chloride, we have to now identify how much is present in the target solution,

V = 1 .2 moles / ( 1.0 moles / L )

V = 1.2 L = 1200 mL

_______________________________________________________

Vadded = 1200 - 500 = 700 mL

Hope that helps!

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Need help with 22 and 24

choli [55]

Answer:

22:

Formular:

$atomic \: mass = \frac{ \sum(isotopic \: mass \times \%abundance)}{100} \\$

substitute:

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23:

Same element is represented by same number of protons. 

Answer:

6 protons. 6 protons

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Note: Atoms with same proton number but different mass number are called isotopes

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3 years ago

Read 2 more answers

After standardizing a NaOH solution, you use it to titrate an HCl solution known to have a concentration of 0.203 M. You perform

jek_recluse [69]

Answer:

0.203 is the mean of the concentration of the HCl solution

Explanation:

You have 5 concentrations. The most appropiate result is the mean of those results. The mean is a statistical defined as the sum of each result divided by the total amount of results. For the results of the problem, the mean is:

0.210 + 0.204 + 0.201 + 0.202 + 0.197 = 1.014 / 5 =

<h3>0.203 is the mean of the concentration of the HCl solution</h3>

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3 years ago

How would you test a colorless crystalline compound to determine if it was a hydrate?

Cloud [144]

A hydrate is a substance where in it contains water and other constituent elements. To know whether if that compound was a hydrate,you should record its mass, then put it in a test tube and heat it with a Bunsen burner. If the compound is a hydrate, the water in the compound will discharge in the form of water vapor. At the next 5-10 minutes, remove it in the test tube and weigh it up again. If the mass is now fewer, that means that there was water existing that has now evaporated, and the compound was a hydrate.

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4 years ago

The leaves of the rhubarb plant contain high concentrations of diprotic oxalic acid (hooccooh) and must be removed before the st

Andrei [34K]

First we have to find Ka1 and Ka2
pKa1 = - log Ka1 so Ka1 = 0.059
pKa2 = - log Ka2 so Ka2 = 6.46 x 10⁻⁵
Looking at the values of equilibrium constants we can see that the first one is really big compared to second one. so, the pH will be affected mainly by the first ionization of the acid.
Oxalic acid is H₂C₂O₄
H₂C₂O₄ ⇄ H⁺ + HC₂O₄⁻
0.0356 M 0 0
0.0356 - x x x
Ka1 = $\frac{[H^+][HC2O4^-]}{[H2C2O4]}$ = x² / 0.0356 - x
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3 years ago

Salt dissolved in water is a solution, therefore _____.

saveliy_v [14]

Answer: Correct options are as follows.

salt is not chemically bonded to water.

salt and water retain their own chemical properties.

Explanation:

When salt is dissolved in water then it means that it is a physical change as salt has completely dissociated into ions but they are not chemically combined to the water molecules.

As a result, both salt and water will retain their chemical properties.

For example, NaCl when dissolved in water will dissociate as follows.

$NaCl \rightarrow Na^{+} + Cl^{-}$

Only the particles of salt have evenly distributed in water.

And, when a components of a salt chemically combine with another substance then it will form a new compound.

Therefore, we can conclude that salt dissolved in water is a solution, therefore:

salt is not chemically bonded to water.

salt and water retain their own chemical properties.

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3 years ago