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slega [8]
2 years ago
11

Consider a triprotic acid like phosphoric acid or citric acid. which expression correctly describes the relative magnitudes of t

he three ka equilibrium constants?
Chemistry
1 answer:
BaLLatris [955]2 years ago
7 0
The triprotic acid like H₃PO₄ contains three protons H⁺ so it ionized through three steps, the acidity strength of first proton higher than second proton higher than third one so Ka1 > Ka2 > Ka3 But
For pKa which equals to -log Ka ... the higher the value of Ka, the smaller the value of pKa. so the correct answer will be:
pKa1 < pKa2 < pKa3
You might be interested in
When the valve between the two containers is opened and the gases allowed to mix, how does the volume occupied by the n2 gas cha
schepotkina [342]
Missing question: 
Nitrogen: <span>2.0 L; </span>1.0 atm; 25°<span>C.
Oxygen: 3</span>.0 L; 2.0 atm; 25°C.
<span>When the valve between the two containers is opened, nitrogen gas moves from one container to another container and gases are mixed together, total volume of nitrogen is than:
V(nitrogen) = 2,0 L + 3,0 L = 5,0 L.</span>
4 0
2 years ago
Read 2 more answers
A 25.0-mL sample of a 1.20 M potassium chloride solution is mixed with 15.0 mL of a 0.900 M lead(II) nitrate solution and this p
xxTIMURxx [149]

Answer:

Limiting reagent = lead(II) nitrate

Theoretical yield = 3.75435 g

% yield = 65.26 %

Explanation:

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For potassium chloride :

Molarity = 1.20 M

Volume = 25.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 25.0×10⁻³ L

Thus, moles of potassium chloride :

Moles=1.20 \times {25.0\times 10^{-3}}\ moles

<u>Moles of potassium chloride  = 0.03 moles</u>

For lead(II) nitrate :

Molarity = 0.900 M

Volume = 15.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 25.0×10⁻³ L

Thus, moles of lead(II) nitrate :

Moles=0.900 \times {15.0\times 10^{-3}}\ moles

<u>Moles of lead(II) nitrate  = 0.0135 moles</u>

According to the given reaction:

2KCl_{(aq)}+Pb(NO_3)_2_{(aq)}\rightarrow PbCl_2_{(s)}+2KNO_3_{(aq)}

2 moles of potassium chloride react with 1 mole of lead(II) nitrate

1 mole of potassium chloride react with 1/2 mole of lead(II) nitrate

0.03 moles potassium chloride react with 0.03/2 mole of lead(II) nitrate

Moles of lead(II) nitrate = 0.015 moles

<u>Limiting reagent is the one which is present in small amount. Thus, lead(II) nitrate is limiting reagent. (0.0135 < 0.015)</u>

The formation of the product is governed by the limiting reagent. So,

1 mole of lead(II) nitrate gives 1 mole of lead(II) chloride

0.0135 mole of lead(II) nitrate gives 0.0135 mole of lead(II) chloride

Molar mass of lead(II) chloride = 278.1 g/mol

Mass of lead(II) chloride = Moles × Molar mass = 0.0135 × 278.1 g = 3.75435 g

<u>Theoretical yield = 3.75435 g</u>

Given experimental yield = 2.45 g

<u>% yield = (Experimental yield / Theoretical yield) × 100 = (2.45/3.75435) × 100 = 65.26 %</u>

6 0
2 years ago
A 2.80 g sample of Al reacts with 4.15 g sample of Cl2 according to the equation shown below.
solong [7]

Answer:

Mass = 5.33 g

Explanation:

Given data:

Mass of Al = 2.80 g

Mass of Cl₂ = 4.15 g

Theoretical yield of AlCl₃ = ?

Solution:

Chemical equation:

2Al  +  3Cl₂        →       2AlCl₃

Number of moles of Al:

Number of moles = mass/molar mass

Number of moles = 2.80 g/ 27 g/mol

Number of moles = 0.10 mol

Number of moles of Cl₂:

Number of moles = mass/molar mass

Number of moles = 4.15 g/71 g/mol

Number of moles = 0.06 mol

Now we will compare the moles of AlCl₃ with Al and Cl₂.

                    Cl₂           :        AlCl₃

                    3              :          2

                   0.06         :        2/3×0.06 = 0.04

                   Al             :        AlCl₃

                     2            :          2

                   0.10         :        0.10

Number of moles of AlCl₃ produced by chlorine are less so it will be limiting reactant.

Mass of AlCl₃:Theoretical yield

Mass = number of moles ×molar mass

Mass = 0.04 mol × 133.34 g/mol

Mass = 5.33 g

4 0
1 year ago
Alkenes: draw the product of 1-chloro-2-ethylcyclohexene with hydrogen gas and a platinum catalyst
atroni [7]
Reacting 1-chloro-2-ethylcyclohexene with hydrogen gas using a platinum catalyst would give a product of 1-chloro-2-ethylcyclohexane. 

Hydrogen gas is a reducing agent, which in this reaction, simply mean that the alkene double bond in the cyclohexene will disappear because one of the two bonds forming the double bond (in the alkene) will be connected to a hydrogen atom. The platinum catalyst is necessary to allow the reaction to proceed at a much lower (activation) energy than would have been required. 

7 0
2 years ago
If water’s density is 1.0 g/mL, then would the perfume be more or less dense than water? Would the perfume float on top or sink
Alex17521 [72]

Answer:

usually the perfumes are made of aromatic hydrocarbons invloving

cetone, ethanol, benzaldehyde, formaldehyde, limonene, methylene chloride, camphor, ethyl acetate, linalool and benzyl alcohol. which have density lower than the water hence they will float on the top of the water.

Hope this helps you

Explanation:

5 0
2 years ago
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