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2 years ago

Why is the sky blue??

Chemistry

1 answer:

borishaifa [10]2 years ago

6 0

Answer:

The gases and particles of earths atmosphere scatter sunlight in different directions making the sky a specific color. Since blue light travels shorter, smaller waves its scattered more than others.

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A weak acid, ha, has a p��a of 4. 524. If a solution of this acid has a ph of 4. 799 , what percentage of the acid is not io

insens350 [35]

Answer:

highly endothermic

Explanation:

6 0

2 years ago

When does phenolphthalein turn pink?

Setler79 [48]

Phenolphthalein turn pink in basic solutions.it turn colourless in acid solution.
Hence options(1)is correct

3 0

3 years ago

Read 2 more answers

Draw the structure of the compound identified by the following simulated 1H and 13C NMR spectra. The molecular formula of the co

Luba_88 [7]

Answer: 4-allylanisole

Explanation: The doublets behind the 7 ppm belongs to the

para-substituted benzene ring. The three single-proton multi-plets around 5−6 ppm predicts that there has to be a single subsituted alkene group

A single plus a doublet around 3-4 ppm belongs to CH3 and CH2 Groups as they could be attached to the subsituted alkene group.

Moreover the interpretation of the NMR that there is no peak with a higher intensity for >180 ppm represents an absence of Carbonyl group.

The Predicted Number is attached from a chemical database along with their peaks information

6 0

3 years ago

What mass of solute is needed to prepare each of the following solutions?

noname [10]

Answer:

a. 21.7824 g

b. 0.2362 g

c. 31.5273 g

Please see the answers in the picture attached below.

Explanation:

Please see the step-by-step solution in the picture attached below.

Hope this answer can help you. Have a nice day!

5 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

2 years ago