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meriva
3 years ago
12

Calculate the percent ionization of propionic acid (C2H5COOH) in solutions of each of the following concentrations (Ka is given

in Appendix D in the textbook).
Part A. 0.270 M

Part B. 7.84×10-2 M

Part C. 1.97×10-2 M
Chemistry
1 answer:
tekilochka [14]3 years ago
7 0

Answer:

a) = 0.704%

b) = 1.30%

c) = 2.60%

Explanation:

Given that:

K_a= 1.34*10^{-5

For Part A; where Concentration of A = 0.270 M

Percentage Ionization(∝)  \alpha = \sqrt{\frac{K_a}{C} }

\alpha = \sqrt{\frac{1.34*10^{-5}}{0.270} }

\alpha = \sqrt{4.9629*10^{-5}}

\alpha = 7.044*10^{-3

percentage% (∝) = 7.044*10^{-3}*100

= 0.704%

For Part B; where Concentration of B = 7.84*10^{-2 M

\alpha = \sqrt{\frac{1.34*10^{-5}}{7.84*10^{-2}} }

\alpha = \sqrt{1.709*10^{-4} }

\alpha = 0.0130\\

percentage% (∝) = 0.0130 × 100%

= 1.30%

For Part C; where Concentration of C= 1.92*10^{-2} M

\alpha = \sqrt{\frac{1.34*10^{-5}}{1.97*10^{-2}} }

\alpha = \sqrt{6.802*10^{-4}}

\alpha =0.02608

percentage% (∝) = 0.02608  × 100%

= 2.60%

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Classify the following reaction according to the five basic reaction types:
joja [24]

Answer:

a. single replacement

Explanation:

Chemical equation:

Cd + H₂SO₄    →    CdSO₄ + H₂

In given reaction Cd replace the hydrogen and form cadmium sulfate and hydrogen gas.

Single replacement:

It is the reaction in which one elements replace the other element in compound.

AB + C → AC + B

Other options are incorrect because,

Combustion:

In combustion reaction substances are burn in the presence of oxygen and form carbon dioxide and water.

Synthesis reaction:

It is the reaction in which two or more simple substance react to give one or more complex product.

A + B  →  AB

Double replacement:

It is the reaction in which two compound exchange their ions and form new compounds.

AB + CD → AC +BD

8 0
3 years ago
Which of the following is a liquid someone help!!!!!
Leya [2.2K]

Answer:

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Explanation:

4 0
3 years ago
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What mass of carbon dioxide is formed when 1.75 mol of ethane burns completely in oxygen?
dybincka [34]

Answer:

There is a mass of 154 Grams of Carbon Dioxide.

Explanation:

One mole is equal to 6.02 × 10^23 particles.

This means we have 1.05 X 10^24 total particles of Ethane.

Each ethane particle contains 2 carbon atoms.

If every particle of ethane is burned, we will end up with 2.10 x 10^24 molecules of Carbon Dioxide (Particles of Methane x 2, since each Methane particle contains 2 carbon atoms)

Carbon Dioxide has a molar mass of 44.01 g/mol

So if we take our amount of Carbon Dioxide molecules and divide it by 1 mole, ((2.10 x 10^24)/(6.02 x 10^23) = 3.49) we find that we have 3.49 moles of Carbon Dioxide.

Now all we need to do is multiply our moles of carbon dioxide(3.49) by it's molar mass(44.01) while accounting for significant digits.

What you should end up with is 154 Grams of Carbon Dioxide.

Hope this helps (And more importantly I hope I didn't make any errors in my math lol)

As a side note this is all assuming that this takes place at STP conditions.

3 0
3 years ago
Calcium bicarbonate (Ca(HCO3)2) Ca:H:C:O = 1:2:2 ___ Lithium sulfide (Li:2s) Li:S = 2:__
Anastaziya [24]

Answer: The ratio of atoms in calcium bicarbonate ; Ca : H : C : O = 1:2:2:6.

The ratio of atoms in lithium sulfide; Li : S = 2 : 1

Explanation:

In calcium bicarbonate:Ca(HCO_3)_2

In a molecular formula of calcium carbonate there are:

Number of Calcium atoms = 1

Number of Hydrogen atom = 1 × 2 = 2

Number of Carbon atoms = 1 × 2 = 2

Number of Oxygen atoms = 3 × 2 = 6

So, Ca : H : C : O = 1 : 2 : 2 : 6

In lithium sulfide :Li_2S

In a molecular formula of lithium sulfide there are:

Number of Lithium atoms =  1 × 2 = 2

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8 0
3 years ago
How many moles of nitrogen are present at STP if the volume is 846L
Artyom0805 [142]
A mole of any gas occupied 22.4 L at STP. So, the number of moles of nitrogen gas at STP in 846 L would be 846/22.4 = 37.8 moles of nitrogen gas.

Alternatively, you can go the long route and use the ideal gas law to solve for the number of moles of nitrogen given STP conditions (273 K and 1.00 atm). From PV = nRT, we can get n = PV/RT. Plugging in our values, and using 0.08206 L•atm/K•mol as our gas constant, R, we get n = (1.00)(846)/(0.08206)(273) = 37.8 moles, which confirms our answer.
3 0
3 years ago
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