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3 years ago

a solution has 23.22% mass/concentration of a particular solvent what is the mass of solute dissolved in 2.5 liter of solvent?

1 answer:

7 0

Let the mass of the solute be x

So, the equation would be
x/2.5+x ×100 = 23.22
x/2.5+x = 23.22/100
100x = 58.05 + 23.22x
100x - 23.22x = 58.05
76.78x = 58.05
x =0.756 ≈ 0.76 litres

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A 50.0 mL solution of 0.141 M KOH is titrated with 0.282 M HCl . Calculate the pH of the solution after the addition of each of

Kobotan [32]

Answer:

pH =1 2.84

Explanation:

First we have to start with the reaction between HCl and KOH:

$HCl~+~KOH->~H_2O~+~KCl$

Now for example, we can use a volume of 10 mL of HCl. So, we can calculate the moles using the molarity equation:

$M=\frac{mol}{L}$

We know that $10mL=0.01L$ and we have the concentration of the HCl $0.282M$ , when we plug the values into the equation we got:

$0.282M=\frac{mol}{0.01L}$

$mol=0.282*0.01$

$mol=0.00282$

We can do the same for the KOH values ( $50mL=0.05L$ and $0.141M$ ).

$0.141M=\frac{mol}{0.05L}$

$mol=0.141*0.05$

$mol=0.00705$

So, we have so far 0.00282 mol of HCl and 0.00705 mol of KOH. If we check the reaction we have a molar ratio 1:1, therefore if we have 0.00282 mol of HCl we will need 0.00282 mol of KOH, so we will have an excess of KOH. This excess can be calculated if we substract the amount of moles:

$0.00705-0.00282=0.00423mol~of~KOH$

Now, if we want to calculate the pH value we will need a concentration, in this case KOH is in excess, so we have to calculate the concentration of KOH. For this, we already have the moles of KOH that remains left, now we need the total volume:

$Total~volume=50mL+10mL=60mL$

$60mL=0.06L$

Now we can calculate the concentration:

$M=\frac{0.00423mol}{0.06L}$

$M=0.0705$

Now, we can calculate the pOH (to calculate the pH), so:

$pOH=-Log(0.0705)$

$pOH=1.15$

Now we can calculate the pH value:

$14=~pH~+~pOH$

$pH=14-1.15=12.84$

8 0

3 years ago

Identify the dominant intermolecular attraction in bh3.

k0ka [10]

The molecule BH3 is trigonal planar, with B in the center and H in the three vertices. Ther are no free electrons. All the valence electrons are paired in and forming bonds.

There are four kind of intermolecular attractions: ionic, hydrogen bonds, polar and dispersion forces.

B and H have very similar electronegativities, Boron's electronegativity is 2.0 and Hydrogen's electronegativity is 2.0.

The basis of ionic compounds are ions and the basis of polar compounds are dipoles.

The very similar electronegativities means that B and H will not form either ions or dipoles. So, that discards the possibility of finding ionic or polar interactions.

Regarding, hydrogen bonds, that only happens when hydrogen bonds to O, N or F atoms. This is not the case, so you are sure that there are not hydrogen bonds.

When this is the case, the only intermolecular force is dispersion interaction, which present in all molecules.

Then, the answer is dispersion interaction.

6 0

3 years ago

describe the" fear appeal" and how it's used in public messaging. Give an example or create your own public health message using

Vsevolod [243]

A good example is cigarretes they try to scare you into not smoking by telling you all of the downsides of smoking like lung cancer and i guess that would be fear appeal

7 0

3 years ago

Which of the following expressions best represents a second order rate law?

Ksivusya [100]

I think the correct answer would be C. The expression that would best represent a second order rate law would be r =k[X][Y]. Reaction with this rate law are those that depend on the concentration of two first order reactants or a second order reactant.

8 0

3 years ago

Suppose a group of volunteers is planning to build a park near a local lake. The lake is known to contain low levels of arsenic

Kisachek [45]

Answer:

A) 10.75 is the concentration of arsenic in the sample in parts per billion .

B) 7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C) It will take 1.37 years to remove all of the arsenic from the lake.

Explanation:

A) Mass of arsenic in lake water sample = 164.5 ng

The ppb is the amount of solute (in micrograms) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation: