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Drupady [299]
2 years ago
11

There is an extreme temperature difference between day and night in a desert during clear weather. However, even though the outs

ide temperature drops greatly at night, the water in a swimming pool in the desert tends to remain warm. Use your knowledge of heat energy, specific heat, and heat transfer to explain these facts.
please help!!!!!! ​
Chemistry
1 answer:
den301095 [7]2 years ago
5 0

Sand has a lower heat capacity hence it heats up easily and looses heat easily. On the other hand, water has a higher heat capacity and looses heat more slowly hence it remains warm at night.

<h3>What is heat transfer?</h3>

The transfer of heat refers to the movement of heat from one body to another. We should note that sand has a low specific heat capacity hence it is rapidly heated up in the day and looses heat quickly at night.

On the other hand, water has a high specific heat capacity hence it heat up slowly and also does not loose heat easily even when the surrounding temperature is much colder. As such, there is an extreme temperature difference between day and night in a desert during clear weather and the water in a swimming pool in the desert tends to remain warm even when the outside temperature drops greatly at night.

Learn more about heat transfer: brainly.com/question/12107378

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A new potential heart medicine, code-named X-281, is being tested by a pharmaceutical company, Pharma-pill. As a research techni
yarga [219]

Answer:

The pKa of X-281 is 3.73.

Explanation:

X-281 is a monoprotic weak acid so that it will not dissociate completely in solution (weak acid) and will only produce 1 mol of protons per mol compound (monoprotic acid).

The dissociation equation could be written as follows:

X-281-H ⇆ X-281 + H⁺

Note the equilibrium arrows indicating that not all X-281-H dissociates at equilibrium.

Initially, the concentration of X-281-H is 0.089 M. At equilibrium, the concentration of the dissociation products, X-281 and H⁺, is unknown but both must be the same, since the drug is a monoprotic acid. We can call "X" to the concentration of the products. The concentration of X-281-H is the initial concentration minus the concentration of the products: 0.089 M - X. Then, at equilibrium, these are the concentrations of each species present:

  X-281-H               ⇆      X-281        +       H⁺

(0.089 M-X)                         X                    X

We also know that the pH is 2.40. Then:

pH = -log[H⁺] = 2.40

where [H⁺] is the molar concentration of the protons or "X". Then:

-log X = 2.40

log X = -2.40

X = 10^(-2.40) = 3.98 x 10⁻³ M

The concentration of each species present in the equilibrium is then:

[H⁺] = 3.98 x 10⁻³ M

[X-281] = 3.98 x 10⁻³ M

[X-281-H] = 0.089 M - 3.98 x 10⁻³M = 0.085 M

At equlibrium, the acidity constant Ka is:

Ka = [X-281] * [H⁺] / [X-281-H]

Ka = (3.98 x 10⁻³ M * 3.98 x 10⁻³ M ) / 0.085 M = 1.86 x 10⁻⁴

Then the pKa is:

pKa = -log Ka = -log (1.86 x 10⁻⁴) = <u>3.73</u>

7 0
4 years ago
The Lewis structure for ethylene, C2H4, is shown. How many valence electrons do the two carbon atoms in the molecule share with
DochEvi [55]
So,

The Lewis structure for ethylene is two carbons double-bonded to each other, and two hydrogens single-bonded to each carbon atom.

Each bond symbolizes two shared valence electrons.  Since the carbon atoms are double-bonded, they share four (4) valence electrons (the other electrons are in lower energy levels and do not appear because they typically don't react when higher energy-level electrons are on top of them).


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3 years ago
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docker41 [41]

Answer:

I think long roots

Explanation:

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Suppose you are performing a titration. at the beginning of the titration, you read the titrant volume as 2.42 ml. after running
Luda [366]

The volume of titrant required for the titration would be 19.81 mL.

Since the burette was not filled to the zero mark during the titration and the level of base titrant was not filled to the 2.42 mL mark. As a result, the difference between the two values represents the total amount of titrant used in the titration.

therefore,

Volume of titrant after running titration - Volume of titrant before running titration  = total titrant required for the titration

22.23 - 2.42 = 19.81 mL

What Exactly Is Titration?

Titration is a laboratory technique that uses a solution with known volume and concentration to determine the concentration of an unknown solution. Between the two solutions, an oxidation-reduction reaction or acid-base neutralization occurs, and the known quantities are used to calculate the unknown. The known concentration standard solution is referred to as the titrant or titrator, while the unknown concentration solution is referred to as the titrand or analyte.

Find more on titration at : brainly.com/question/21881827

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5 0
2 years ago
The reaction was studied at a series of different temperatures. A plot of ln(k) vs. 1/T gave a straight line relationship with a
borishaifa [10]

Here is the correct question:

The reaction A → products was studied at a series of different temperatures. A plot of ln(k) vs. 1/T gave a straight line relationship with a slope of -693 and a y-intercept of -0.425. Additionally, a study of the concentration of A with respect to time showed that only a plot of 1/[A] vs. time gave a straight line relationship. What is the initial rate of this reaction when [A] = 0.41 at 271 K ?

Answer:

the initial rate of this reaction is 0.0216275  M/sec

Explanation:

Using the formula:

K = Ae^{\frac{-Ea}{RT}}

InK = InA + (\frac{-Ea}{R})(\frac{1}{T})\\y  \ \ \ \ \ \ \  \ \ \ c \ \ \ \ \ \ \ \ \ \ m  \ \ \ \ x

m = \frac{-Ea}{R} = -693

\frac{Ea}{8.314}= 693 \ \\ \\ Ea = 693 * 8.314 \\ \\ Ea = 5671.602 \ J

In A = -0.425 \ \ \\ \\  A = e^{-0.425} \\   \\  A = 0.6538

K = 0.6538 e^{- (\frac{5761.602}{8.314*271})

K = 0.05275 \\ \\ K = 5.275*10^{-2}

Since \frac{1}{[A]} vs time is a straight line relationship;

Therefore, it is a second order reaction

rate = K[A]²

rate = 5.275 × 10⁻² × (0.41)

rate = 0.0216275  M/sec

8 0
3 years ago
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