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Anit [1.1K]
2 years ago
6

Study Example 8.1 on page 321. Then, calculate the molarity of a solution made by adding 5.51 g of CaCl2 to enough water to crea

te a solution with a total volume of 2.50 L. Enter only a numerical value in the correct number of significant figures. Use the Periodic Table that is part of your exam to calculate molar mass. Do not enter units in the answer box, but know that the symbol for the unit of this calculation is M
Chemistry
1 answer:
DiKsa [7]2 years ago
7 0

Answer:

0.0198 M

Explanation:

Given data

  • Mass of CaCl₂ (solute): 5.51 g
  • Volume of solution: 2.50 L

Step 1: Calculate the moles of solute

The <em>molar mass of CaCl₂</em> is 110.98 g/mol. The moles corresponding to 5.51 g are:

5.51 g × (1 mol/110.98 g) = 0.0496 mol

Step 2: Calculate the molarity of the solution

<em>M = moles of solute / liters of solution</em>

M = 0.0496 mol / 2.50 L

M = 0.0198 M

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A solution of SO2 in water contains 0.00023 g of SO2 per liter of solution. What is the concentration of SO2 in ppm? in ppb?
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Answer:

= 230 ppb

Explanation:

Considering that;

1ppm = 1mg/L  

Then;

0.00023g = 0.23mg  

Therefore;

0.00023 g/L = 0.23 mg/L

0.23 mg/L = 0.23 ppm

1 ppm = 100 ppb

Therefore;

0.23 ppm = 0.23 ×1000

                = 230 ppb

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25.0 ml of a 6.0 m hno3 stock solution is diluted using water to 100 ml. How many moles of hno3 are present in the dilute soluti
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Answer:

The answer to your question is: 6 moles of HNO₃

Explanation:

Data

Volume = 25 ml

Concentration = 6 M HNO₃

Diluted 100 ml

Formula

Molarity = # moles / volume

# of moles = Volume x Molarity

Process

# of moles = 0.10 x 6

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The oxidation numbers of nitrogen in NH3, HNO3, and NO2 are, respectively: -3, -5, +4 +3, +5, +4 -3, +5, -4 -3, +5, +4
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In NH3 , let oxidation number of N be x

x + (+1)3 = 0

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1 + x + (-2)3 = 0

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Which of the compounds above are strong enough acids to react almost completely with a hydroxide ion (pka of h2o = 15.74) or wit
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The compounds can react with OH⁻ and HCO₃⁻ only C₅H₆N pyridinium

<h3><em>Further explanation </em></h3>

In an acid-base reaction, it can be determined whether or not a reaction occurs by knowing the value of pKa or Ka from acid and conjugate acid (acid from the reaction)

Acids and bases according to Bronsted-Lowry

Acid = donor (donor) proton (H + ion)

Base = proton (receiver) acceptor (H + ion)

If the acid gives (H +), then the remaining acid is a conjugate base because it accepts protons. Conversely, if a base receives (H +), then the base formed can release protons and is called the conjugate acid from the original base.

From this, it can be seen whether the acid in the product can give its proton to a base (or acid which has a lower Ka value) so that the reaction can go to the right to produce the product.

The step that needs to be done is to know the pKa value of the two acids (one on the left side and one on the right side of the arrow), then just determine the value of the equilibrium constant

Can be formulated:

K acid-base reaction = Ka acid on the left : K acid on the right.

or:

pK = acid pKa on the left - pKa acid on the right

K = equilibrium constant for acid-base reactions

pK = -log K;

K~=~10^{-pK}

K value> 1 indicates the reaction can take place, or the position of equilibrium to the right.

There is some data that we need to complete from the problem above, which is the pKa value of some compounds that will react, namely:

pyridinium pKa = 5.25

acetone pKa = 19.3

butan-2-one pKa = 19

Let's look at the K value of each possible reaction:

pka H₂O = 15.74, pka of H₂CO₃ = 6.37)

  • 1. C₅H₆N pyridinium

* with OH⁻

C₅H₆N + OH- ---> C₅H₅N- + H₂O

pK = pKa pyridinium - pKa H₂O

pK = 5.25 - 15.74

pK = -10.49

K~=~10^{4.9}

K values> 1 indicate the reaction can take place

* with HCO3⁻

C₅H₆N + HCO₃⁻-- ---> C₅H₅N⁻ + H₂CO₃

pK = 5.25 - 6.37

pK = -1.12

K`=~10^{1.12]

Reaction can take place

  • 2. Acetone C₃H₆O

* with OH-

C₃H₆O + OH⁻ ---> C₃H₅O- + H₂O

pK = 19.3 - 15.74

pK = 3.56

K~=~10^{ -3.56}

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* with HCO₃-

C₃H₆O + HCO₃⁻ ----> C₃H₅O⁻ + H₂CO₃

pK = 19.3 - 6.37

pK = 12.93

K`=~10 ^{-12.93}

Reaction does not happen

  • 3. butan-2-one C₄H₇O

* with OH-

C₄H₇O + OH- ---> C₄H₆O- + H₂O

pK = 19 - 15.74

pK = 3.26

K~=~10^{-3.26}

Reaction does not happen

* with HCO₃⁻

C₄H₇O + HCO₃⁻ ---> C₄H₆O⁻ + H₂CO₃

pK = 19 - 6.37

pK = 12.63

K~=~ 10^{-12.63}

Reaction does not happen

So that can react with OH⁻ and HCO₃⁻ only C₅H₆N pyridinium

<h3><em>Learn more </em></h3>

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