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3 years ago

7

life would not be possible on earth without the water that covers much of its surface, and the air that surrounds it. earrhs wat

er and air affect each other in many ways. which of the following situations represents an interaction between the atmosphere and the hydrosphere

1 answer:

maks197457 [2]3 years ago

4 0

Answer:evaporation from lakes and rivers

Explanation:I took the test

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How many quarters fit in a 1.75 liter bottle?

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1.85 quarts can fit into a 1.75 liter bottle

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Acetaldehyde (CH3CHO) undergoes a Wolf-Kishner reaction, which is the addition of hydrazine (H2NNH2) with subsequent addition of

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Reduced, alkane

Explanation:

Wolf-Kishner reaction is a type of reduction reaction in which aldehydes and ketones are reduced to their corresponding alkane in the presence of a base.

This reaction occurs at high temperature.

Alkane formed has a same number of carbon as aldehyde and ketone.

If acetaldehyde undergoes a Wolf-Kishner reaction in the presence of base and heat, then ethane is formed. Nitrogen is formed as a byproduct.

Here, acetaldehyde is reduced to form ethane.

So, acetaldehyde undergoes a Wolf-Kishner reaction, which is the addition of hydrazine with subsequent addition of a base and heat. In this reaction, the aldehyde is reduced, resulting in alkane product.

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3 years ago

How many grams are in 4 moles of aluminum phosphate

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Name two elements that have the same properties as calcium (Ca). (4 points)

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3 years ago

Dinitrogen tetraoxide, N2O4, decomposes to nitrogen dioxide, NO2, in a first-order process. If k = 1.5 x 103 s-1 at 5 ºC and k =

Arada [10]

Answer:

The activation energy for the decomposition = 33813.28 J/mol

Explanation:

Using the expression,

$\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )$

Wherem

$k_1\ is\ the\ rate\ constant\ at\ T_1$

$k_2\ is\ the\ rate\ constant\ at\ T_2$

$E_a$ is the activation energy

R is Gas constant having value = 8.314 J / K mol

Thus, given that, $E_a$ = ?

$k_2=4.0\times 10^3s^{-1}$

$k_1=1.5\times 10^3s^{-1}$

$T_1=5\ ^0C$

$T_2=25\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (5 + 273.15) K = 278.15 K

T = (25 + 273.15) K = 298.15 K

$T_1=278.15\ K$

$T_2=298.15\ K$

So,

$\ln \frac{1.5\times 10^3}{4.0\times 10^3}\:=-\frac{E_{a}}{8.314}\times \left(\frac{1}{278.15}-\frac{1}{298.15}\right)$

$E_a=-\ln \frac{1.5\times \:10^3}{4.0\times \:10^3}\:\times \frac{8.314}{\left(\frac{1}{278.15}-\frac{1}{298.15}\right)}$

$E_a=-\frac{8.314\ln \left(\frac{1.5\times \:10^3}{4\times \:10^3}\right)}{\frac{1}{278.15}-\frac{1}{298.15}}$

$E_a=-\frac{689483.53266 \ln \left(\frac{1.5}{4}\right)}{20}$

$E_a=33813.28\ J/mol$

<u>The activation energy for the decomposition = 33813.28 J/mol</u>

8 0

3 years ago