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katen-ka-za [31]
3 years ago
5

Which factor MOST directly affects the flow of ocean currents?w of ocean currents?

Chemistry
1 answer:
Lynna [10]3 years ago
6 0

B)  differences in temperature

Explanation:

The factor that most directly affects the flow of ocean currents is the differences in ocean temperature from one place to another.

Ocean current is the directional movement of the ocean water in its basin.

The main cause of ocean current is differences in temperature.

  • Ocean water on the surface receives direct sunlight.
  • This warms and the surface and decreases their density.
  • When winds blows over them, they cool and sink under their own weight.
  • The less cold water in depths rise to replace them.
  • This exchange of ocean water due to loss and gain of heat cause ocean current.

learn more:

Ocean density brainly.com/question/6760255

#learnwithBrainly

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What happens to energy when methane increases?
Zanzabum

Answer:

How does methane gas affect the environment?

If methane leaks into the air before being used — from a leaky pipe, for instance — it absorbs the sun's heat, warming the atmosphere. For this reason, it's considered a greenhouse gas, like carbon dioxide.

Explanation:

4 0
3 years ago
2) A common "rule of thumb" -- for many reactions around room temperature is that the
babunello [35]

The question is incomplete. The complete question is :

A common "rule of thumb" for many reactions around room temperature is that the rate will double for each ten degree increase in temperature. Does the reaction you have studied seem to obey this rule? (Hint: Use your activation energy to calculate the ratio of rate constants at 300 and 310 Kelvin.)

Solutions :

If we consider the activation energy to be constant for the increase in 10 K temperature. (i.e. 300 K → 310 K), then the rate of the reaction will increase. This happens because of the change in the rate constant that leads to the change in overall rate of reaction.

Let's take :

$T_1=300 \ K$

$T_2=310 \ K$

The rate constant = $K_1 \text{ and } K_2$ respectively.

The activation energy and the Arhenius factor is same.

So by the arhenius equation,

$K_1 = Ae^{-\frac{E_a}{RT_1}}$  and $K_2 = Ae^{-\frac{E_a}{RT_2}}$

$\Rightarrow \frac{K_1}{K_2}= \frac{e^{-\frac{E_a}{RT_1}}}{e^{-\frac{E_a}{RT_2}}} $

$\Rightarrow \frac{K_1}{K_2}=  e^{-\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}$

$\Rightarrow \ln \frac{K_1}{K_2}= - \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}=  \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

Given, $E_a = 0.269$ J/mol

           R = 8.314 J/mol/K

$\Rightarrow \ln \frac{K_2}{K_1}=  \frac{0.269}{8.314} \left(\frac{1}{300} -\frac{1}{310} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}=  \frac{0.269}{8.314} \times \frac{10}{300 \times 310}$

$\Rightarrow \ln \frac{K_2}{K_1}=  3.479 \times 10^{-6}$

$\Rightarrow  \frac{K_2}{K_1}=  e^{3.479 \times 10^{-6}}$

$\Rightarrow  \frac{K_2}{K_1}=  1$

∴ $K_2=K_1$

So, no this reaction does not seem to follow the thumb rule as its activation energy is very low.

8 0
2 years ago
If a patient\'s blood pressure is 145 over 65 mmHg, what is it in atmospheres (atm)
kicyunya [14]

Converting mmHg to atm is solved by division.

Example: Convert 745.0 to atm.

Solution- divide the mmHg value by the 760.0 mmHg / atm.

745 mmHg over 760.0 mmHg/atm

atm value is 0.980263

Now, I am a medical student and we have never had to convert a BP (blood pressure) to atm from mmHg, only ever kPA. SO, I am going to take a guess here and say that when you do the work to solve this, you are going to convert the Systolic (upper #) which is the 145. You should get 0.190789 and then convert the Diastolic (lower #) which is 65. You should get 0.08552632.

So your fraction so to speak should read, 0.190789/0.08552632 or 0.190789 over 0.08552632

(Just to note that is way to low of a BP, although it is irrelevant) Best wishes and good luck. "Remember, never just look for the right answer, look for why it is the right answer!"

7 0
3 years ago
Read 2 more answers
How to prepare magnesium carbonate starting with magnesium nitrate
STALIN [3.7K]

Answer:

How to prepare Magnesium Carbonate:

Explanation:

Magnesium carbonate can be prepared in laboratory by reaction between any soluble magnesium salt and sodium bicarbonate: MgCl2(aq) + 2NaHCO3(aq) → MgCO3(s) + 2NaCl(aq) + H2O(l) + CO2(g)

8 0
3 years ago
HELP! <br> A. W<br> B. X<br> C. Y<br> D. Z
tamaranim1 [39]

Answer:

B

Explanation:

7 0
2 years ago
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