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stira [4]
2 years ago
10

How do weather models predict the weather?

Chemistry
1 answer:
strojnjashka [21]2 years ago
4 0

Answer:

<em>Weather forecasters often discuss the models they use to help predict the weather. ... Weather observations (pressure, wind, temperature and moisture) obtained from ground sensors and weather satellites are fed into these equations. The observations are brought into the models in a process known as data assimilation.</em>

Explanation:

<h2><em>I </em><em>hope</em><em> this</em><em> helps</em><em>!</em></h2>
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Base your answer on the information below. The hydrocarbon 2-methylpropane reacts with iodine as represented by the balanced equ
Kryger [21]

Answer:

See explanation

Explanation:

The boiling point of a substance is affected by the nature of bonding in the molecule as well as the nature of intermolecular forces between molecules of the substance.

2-methylpropane has only pure covalent and nonpolar C-C and C-H bonds. As a result of this, the molecule is nonpolar and the only intermolecular forces present are weak dispersion forces. Therefore, 2-methylpropane has a very low boiling point.

As for 2-iodo-2-methylpropane, there is a polar C-I bond. This now implies that the intermolecular forces present are both dispersion forces and dipole interaction. As a result of the presence of stronger dipole interaction between 2-iodo-2-methylpropane molecules, the compound has a higher boiling point than  2-methylpropane.

3 0
3 years ago
What is significant about 42?​
maks197457 [2]

Answer:

It’s twice as much as 9+10

Explanation:

5 0
3 years ago
What happens in a condensation reaction?
Agata [3.3K]

A) Head to tail joining of monomers. :) (confirmed correct answer, I took the test)

8 0
3 years ago
The reaction C 4 H 8 ( g ) ⟶ 2 C 2 H 4 ( g ) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ / mol. 262 kJ/mol. At 600.0 K,
ludmilkaskok [199]

Answer: 4.3\times 10^{-13}s^{-1}

Explanation:

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

or,

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = rate constant at 600.0K = 6.1\times 10^{-8}s^{-1}

K_2 = rate constant at 775.0 = ?

Ea = activation energy for the reaction = 262 kJ/mol = 262000J/mol

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 600.0K

T_2 = final temperature = 775.0K

Now put all the given values in this formula, we get

\log (\frac{6.1\times 10^{-8}}{K_2})=\frac{262000}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{775.0K}]

\log (\frac{6.1\times 10^{-8}s^}{K_2})=5.150

(\frac{6.1\times 10^{-8}}{K_2})=141253.8

Therefore, the value of the rate constant at 775.0 K is 4.3\times 10^{-13}s^{-1}

5 0
3 years ago
How do you solve this ?NH4+NO4-&gt;N2+O2+H2O2
ololo11 [35]
I'm assuming you wanted this equation balanced? If you typed it correctly, it's already balanced for you. :)
8 0
3 years ago
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