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BartSMP [9]
3 years ago
7

Please write thoughtful, concise summaries of your findings in paragraph form. Include key calculation values and point out what

those values mean for the specific energy technology. 39 POINTSSSSS
Chemistry
1 answer:
dedylja [7]3 years ago
7 0
I am not sure how to do this
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Gennadij [26K]

Newton was the scientist

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2 years ago
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Classify each of the following compounds as a strong acid, weak acid, strong base, or weak base, and write the Ka expression for
Katyanochek1 [597]

Answer:

See explanation below

Explanation:

There are several ways to know if an acid or base is strong. One method is calculating the pH. If the pH is really low, is a strong acid, and if it's really high is a strong base.

However we do not have a pH value here.

The other method is using bronsted - lowry theory. If an acid is strong, then his conjugate base is weak. Same thing with the bases.

Now, Looking at the 4 compounds, we can say that only two of them is weak and the other two are strong compounds. Let's see:

LiOH ---> Strong. If you try to dissociate :

LiOH ------> Li⁺ + OH⁻     The Li⁺ is a weak conjugate acid.

HF -----> Weak

HF --------> H⁺ + F⁻   The Fluorine is a relatively strong conjugate base.

HCl -----> Strong

This is actually one of the strongest acid.

NH₃ ------> Weak

Now writting the Ka and Kb expressions:

Ka = [H⁺] [F⁻] / [HF]

Kb = [NH₄⁺] [OH⁻] / [NH₃]

Finally, to calculate the [OH⁻] we need to use the following expression:

Kw = [H⁻] [OH⁻]

Solving for [OH⁻] we have:

[OH⁻] = Kw / [H⁺]

Remember that the value of Kw is 1x10⁻¹⁴. So replacing:

[OH⁻] = 1x10⁻¹⁴ / 7x10⁻⁶

[OH⁻] = 1.43x10⁻⁹ M

And now, multiplying by 10¹⁰ we have:

[OH⁻] = 1.429x10⁻⁹ * 1x10¹⁰

<h2>[OH⁻] = 14.29 </h2>

Hope this helps

4 0
2 years ago
Choose all that apply.
Mashutka [201]
A and D would be correct
5 0
2 years ago
How to determine the relative rates of reactant disappearance and product appearance using reaction stoichiometry
Lady_Fox [76]

Answer:

For example, the relative rate of a reaction at 20 seconds will be 1/20 or 0.05 s -1, while the average rate of reaction over the first 20 seconds will be the change in mass over that period, divided by the change in time.

Explanation:

Hope it helps u

FOLLOW MY ACCOUNT PLS PLS

6 0
3 years ago
The chemical equation below is correctly balanced.
DIA [1.3K]

Answer:

44.8 L of O2 will react (option D)

Explanation:

Step 1: Data given

Number of moles of SO2 = 4.00 moles

STP = Pressure = 1 atm  and temperature = 273 K

Step 2: The balanced equation

2 SO2(g) + O2(g) → 2 SO3(g)

Step 3: Calculate moles of O2

For 2 moles SO2, we need 1 mol O2 to produce 2 moles SO3

For 4.00 moles SO2 we need 4.00 / 2 = 2.00 moles O2

Step 4: Calculate volume of O2

For 1 mol we have a volume of 22.4 L

V = (n*R*T)/ p

V = (2.00 * 0.08206 * 273)/p

V = 44.8 L

For 2.00 moles we have a volume of 2*22.4 = 44.8 L

44.8 L of O2 will react (option D)

8 0
3 years ago
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