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harkovskaia [24]
3 years ago
11

Please answer! will mark brainiest for correct answer!

Chemistry
2 answers:
Debora [2.8K]3 years ago
3 0

Answer:

when the electron moves from the first orbit to the third it gains energy  but when it jumps from a higher energy orbit to a lower the energy is being radiated. So whihever is the higher energy you know what it is.

Explanation:

7nadin3 [17]3 years ago
3 0

Answer: So  ever is the higher energy you know what it is.

Explanation:

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How many kilojoules are required to heat 0.500 g of water from 20.3 degrees Celcius to 29.7 degrees Celcius ? How many Calories?
Molodets [167]

Amount of heat = mass of water * specific heat (temperature change

)

= 0.500 g * 4.184 J / g-C ( 29.7 -20.3 )C

= 19.6648 J  

= 0.0197 KJ

And  

1 cal = 4.186798 J

19.6648 J * 1 cal / 4.186798 J =4.70 cal


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3 years ago
Which excerpt from "Bone Detective" best shows that Diane France is recognized as an expert in her line of work?
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Answer:

It is the first one

Explanation:

I already took the quiz so first me I also got a 100% in it and I passed college

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The placement of carbonyl group of a kerose sugar is at second-carbon only (b) fir + carbon only () first and fast carbon (d) li
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Answer:

(d)

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The addictive power of nicotine is enhanced by which of the following chemicals?
ArbitrLikvidat [17]
The correct answer is A. Ammonia.
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4 years ago
How many grams of sodium benzoate should be added to prepare the buffer? Neglect the small volume change that occurs when the so
expeople1 [14]

Answer:

2.72 grams

Explanation:

Complete question is

You are asked to pre pare a pH = 4.00 buffer starting from 1.50 L of 0.0200 M solution of benzoic acid

(C6H5COOH)

and any amount you need of sodium benzoate

(C6H5COONa)

How many grams of sodium benzoate should be added to prepare the buffer? Neglect the small volume change that occurs when the sodium benzoate is added.

Solution

Given

pH of the buffer solution = 4

Concentration of C6H5COOH = 0.02 M

Volume of the buffer solution = 1.50 L

K_a value for benzoic acid is 6.3 * 10^ {-5}

Concentration of sodium benzoate

pH = - log Ka + log \frac{C6H5COONa}{C6H5COOH}

Substituting the given values we get

log \frac{C6H5COONa}{C6H5COOH}  = 4 + log (6.3 * 10^ {-5})\\log \frac{C6H5COONa}{C6H5COOH}  = -0.20\\\frac{C6H5COONa}{C6H5COOH} = 10^{-0.2}\\{C6H5COONa} = 10^{-0.2} * 0.02\\{C6H5COONa} =  0.63 * 0.02\\{C6H5COONa} =  0.0126 M

Number of moles in sodium benzoate

= 0.0126 * 1.5 \\= 0.0189 Mol

Mass of sodium benzoate

0.0189 mol * 144.147 \frac{g}{mol} \\= 2.72 g

3 0
3 years ago
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