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Luba_88 [7]
3 years ago
11

The energy exchanges between space, the atmosphere, and Earth’s surface produce what

Chemistry
2 answers:
nekit [7.7K]3 years ago
8 0

Answer: It produces climate or weather.

Explanation: The energy exchanges between space, the atmosphere, and the Earth's surface produces climate or weather which is responsible for the various life forms that exists on the earth. can be defined as the day to day changes in the temperature and precipitation. Whereas climate refers to the the average changes in atmospheric conditions for a longer period of time.

Sauron [17]3 years ago
7 0
The energy exchanges between space,the atmosphere and the Earth surface produce Weather and Climate
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nikitadnepr [17]

Answer:

Mass = 14.3 g

Explanation:

Given data:

Mass of Mg(OH)₂  = 16.0 g

Mass of HCl = 11.0 g

Mass of MgCl₂ = ?

Solution:

Chemical equation:

Mg(OH)₂ + 2HCl    →   MgCl₂ + 2H₂O

Number of moles of Mg(OH)₂ :

Number of moles = mass/ molar mass

Number of moles = 16.0 g/ 58.3 g/mol

Number of moles = 0.274 mol

Number of moles of HCl :

Number of moles = mass/ molar mass

Number of moles = 11.0 g/ 36.5 g/mol

Number of moles = 0.301 mol

Now we will compare the moles of Mg(OH)₂  and HCl with MgCl₂.

                           Mg(OH)₂          :           MgCl₂

                                 1                 :               1

                                 0.274        :          0.274

                               HCl             :              MgCl₂

                                  2              :               1

                                0.301         :           1/2×0.301 = 0.150

The  number of moles of MgCl₂ produced by HCl are less so it will limiting reactant.

Mass of MgCl₂:

Mass = number of moles × molar mass

Mass = 0.150 ×  95 g/mol

Mass = 14.3 g

8 0
3 years ago
Which prey is common to all owls?
IrinaK [193]

Answer:

Mice

Explanation:

Small, rodent-like mammals, such as voles and mice, are the primary prey for many owl species. An owl's diet may also include frogs, lizards, snakes, fish, mice, rabbits, birds, squirrels, and other creatures. Occasionally, Great Horned Owls might even find skunks tasty enough to eat.

Please mark brainliest! I only need one more to rank up! Thank you and have a blessed day/night :D

8 0
2 years ago
Read 2 more answers
What atom is the smallest size? A. Li B. Be C. F D. C
s2008m [1.1K]

Answer:

A

Explanation:

lies to the left of periodic table

7 0
3 years ago
Help me, please. I Don't get my homework.
Lisa [10]
We’d have to be very careful because if we had our skeletons on the outside it’d be very easy to injure ourselves
7 0
3 years ago
Read 2 more answers
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
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