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djverab [1.8K]
3 years ago
7

Based on the passage , where in the United States would a physician encounter the most cases of SAD ?

Chemistry
1 answer:
Montano1993 [528]3 years ago
7 0

Answer/Explanation:

We cannot see the passage, but SAD stands for Seasonal Affective Disorder. SAD is a type of depressive disorder related to lack of light, and therefore changes with the seasons. It is a severe version of the "winter blues"

In the winter months, where the days are shorter, individuals who suffer from SAD suffer from mood changes. They get depressed, have a reduction in energy levels, and withdraw socially. They may also gain weight. This causes problems with relationships, work, and school.

Since SAD is linked to light, SAD is likely more common in the Northern areas of the United States, where winter days are shorter.

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How does Earth's rotation affect ocean currents?
Tom [10]

Answer:

Our planet's rotation produces a force on all bodies moving relative to theEarth. ... The force, called the "Coriolis effect," causes the direction of winds and ocean currents to be deflected.

Explanation:

4 0
2 years ago
Explain the most important difference between a physical and chemical change.
timama [110]
You can see a physical change always but not always a chemical
7 0
3 years ago
Read 2 more answers
"acid is responsible for the odor in rancid butter. a solution of 0.25 m butyric acid has a ph of 2.71. what is the ka for"
Salsk061 [2.6K]

Answer:- The Ka for the acid is 1.53*10^-^5 .

Solution:- In general, monoprotic acid could be represented by HA. The dissociation equation for the ionization of HA is written as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

Now, we make the ice table for this equation as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

I 0.25 0 0

C -X +X +X

E (0.25 - X) X X

where, I stands for initial concentration, C stands for change in concentration and E stands for equilibrium concentration.

X is the change in concentration and from ice table it's same as the concentration of hydrogen ion that is calculated from given pH.

Ka = [H^+][A^-]\frac{1}{HA}

Where, Ka is the acid ionization constant. Let's plug in the values.

Ka = \frac{X^2}{0.25-X}

Let's calculate the value of X first using the equation:

pH = -log[H^+][/tex]

on taking antilog ob above equation we get:

[H^+]=10^-^p^H

[H^+]=10^-^2^.^7^1

[H^+] = 0.00195

So, X = 0.001195

Let's plug in this value of X in the equation:-

Ka=\frac{(0.00195)^2}{0.25-0.00195}

Ka=1.53*10^-^5

So, the value of Ka for butyric acid is 1.53*10^-^5 .

8 0
3 years ago
Read 2 more answers
Calculate the ph of the solution made by adding 0.50 mol of hobr and 0.30 mol of kobr to 1.00 l of water. the value of ka for ho
nikdorinn [45]
To fully understand the problem, we use the ICE table to identify the concentration of the species. We calculate as follows:

Ka = 2.0 x 10^-9 = [H+][OBr-] / [HOBr] 

HOBr  = 0.50 M 
KOBr  = 0.30 M = OBr-

<span>     HOBr + H2O <-> H+ + OBr- </span>
<span>I     0.50        -              0          0.30 </span>
<span>C       -x                        x             x 
</span>---------------------------------------------
<span>E(0.50-x)                    x       (0.30+x) </span>

<span>Assuming that the value of  x is small as compared to 0.30 and 0.50 </span>

<span>Ka = 2.0 x 10^-9 = x (0.30) / 0.50) </span>

<span>x = 3.33 x 10^-9 = H+</span>
pH = 8.48
3 0
3 years ago
Part 1: What is the final volume in milliliters when 0.730 L of a 44.8 % (m/v) solution is diluted to 23.3 % (m/v)?
Andre45 [30]

part 1 : the final volume : 1.404 L

part 2 : the initial concentration : 4.06 M

<h3>Further explanation </h3>

Dilution is the process of adding a solvent to get a more dilute solution.

The moles(n) before and after dilution are the same.

Can be formulated :

M₁V₁=M₂V₂

M₁ = Molarity of the solution before dilution  

V₁ = volume of the solution before dilution  

M₂ = Molarity of the solution after dilution  

V₂ = Molarity volume of the solution after dilution

part 1 :

M₁=44.8%

V₁=0.73 L

M₂=23.3%

\tt V_2=\dfrac{M_1.V_1}{M_2}\\\\V_2=\dfrac{44.8\times 0.73}{23.3}\\\\V_2=1.404~L

part 2 :

V₁=739 ml=0.739 L

V₂=1.5 L

M₂=2

\tt M_1=\dfrac{M_2.V_2}{V_1}\\\\M_1=\dfrac{2\times 1.5}{0.739}\\\\M_1=4.06

6 0
3 years ago
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