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2 years ago

HELP ASAP! I will give brainiest if you answer all questions correctly.

Chemistry

2 answers:

oksano4ka [1.4K]2 years ago

4 0

Answer:

28) -88kj

29) 300kj

30) -204kj

DaniilM [7]2 years ago

4 0

Answer:

28). 261

29). 788

30).-6201

Explanation:

I just took lesson and it said those were correct, I added screenshots of the lesson if you need it :)

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Calculate the pH of a buffer solution prepared by dissolving 0.20 mole of cyanic acid (HCNO) and 0.80 mole of sodium cyanate (Na

Afina-wow [57]

The pH of a buffer solution : 4.3

<h3>Further explanation</h3>

Given

0.2 mole HCNO

0.8 mole NaCNO

1 L solution

Required

pH buffer

Solution

Acid buffer solutions consist of weak acids HCNO and their salts NaCNO.

$\tt \displaystyle [H^+]=Ka\times\frac{mole\:weak\:acid}{mole\:salt\times valence}$

valence according to the amount of salt anion

Input the value :

$\tt \displaystyle [H^+]=2.10^{-4}\times\frac{0.2}{0.8\times 1}\\\\(H^+]=5\times 10^{-5}\\\\pH=5-log~5\\\\pH=4.3$

7 0

2 years ago

What is the kinetic energy acquired by the electron in hydrogen atom, if it absorbs a light radiation of energy 1.08x101 J. (A)

Delvig [45]

Explanation:

The given data is as follows.

Energy of radiation absorbed by the electron in hydrogen atom = $1.08 \times 10^{-17} J$

As energy is absorbed as a photon. Hence, frequency will be calculated will be as follows.

E = $h \nu$

$1.08 \times 10^{-17} J$ = $6.626 \times 10^{-34} Js \times \nu$

$\nu$ = $0.163 \times 10^{17} s^{-1}$

or, $\nu$ = $1.63 \times 10^{16} s^{-1}$

It is known that, $\nu = \frac{c}{\lambda}$

$1.63 \times 10^{16} s^{-1} = \frac{3 \times 10^{8} m/s}{\lambda}$

$\lambda$ = $1.84 \times 10^{-8} m$

And, according to De-Broglie equation $\lambda = \frac{h}{p}$

as, p = $m \times \nu$

So, $\lambda = \frac{h}{m \times \nu}$

$m \times \nu = \frac{6.626 \times 10^{-34} Js}{1.84 \times 10^{-8} m}$

= $3.6 \times 10^{-26} J/m$

Now, on squaring both the sides we get the following.

$(m \times \nu)^{2}$ = $(3.6 \times 10^{-26} J/m)^{2}$

= $12.96 \times 10^{-52}$

$m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}$

where, m = mass of electron

So, $m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}$

= $\frac{12.96 \times 10^{-52}}{9.1 \times 10^{-31}}$

= $1.42 \times 10^{-21}$ J

Since, K.E = $\frac{1}{2}m \nu^{2}$

= $\frac{1.42 \times 10^{-21} J}{2}$

= $0.71 \times 10^{-21} J$

Thus, we can conclude that kinetic energy acquired by the electron in hydrogen atom is $7.1 \times 10^{-22} J$ .

4 0

2 years ago

A sample of ammonia has a mass of 82.9 g. how many molecules are in this sample?

alina1380 [7]

The chemical formula for ammonia is NH3. So first, you need to find the molar mass of ammonia (how many grams in one mole).
N=14g
H3=3g
So one mole of NH3 is 17 grams, you can divide 82.9 grams by 17 grams to find the number of molecules. The answer should be 4.876 moles (molecules) of ammonia. Hope this helps!

8 0

2 years ago

Read 2 more answers

PLS HELP! 50 PTS!

ioda

Answer:

The average kinetic energy of the gas particles is greater in container B because it has a higher temperature.

Explanation:

<em>The correct option would be that the average kinetic energy of the gas particles is greater in container B because it has a higher temperature.</em>

<u>According to the kinetic theory of matter, the temperate of a substance is a measure of the average kinetic energy of the molecules of substance. In other words, the higher the temperature of a substance, the higher the average kinetic energy of the molecules of the substance.</u>

In the illustration, the gas in container B showed a higher temperature than that of container A as indicated on the thermometer, it thus means that the average kinetic energy of the molecules of gas B is higher than those of gas A.

5 0

2 years ago

In a study of the decomposition of the compound XX via the reaction

Leya [2.2K]

Answer:

( About ) 0.03232 M

Explanation:

Based on the units for this reaction it should be a second order reaction, and hence you would apply the integrated rate law equation "1 / [X] = kt + 1 / [ $X_o$ ]"

This formula would be true for the following information -

{ $X_o$ = the initial concentration of X, k = rate constant, [ X ] = the concentration after a certain time ( which is what you need to determine ), and t = time in minutes }

________

Therefore, all we have left to do is plug in the known values. The initial concentration of X is 0.467 at a time of 0 minutes, as you can tell from the given data. This is not relevant to the time needed in the formula, as we need to calculate the concentration of X after 18 minutes ( time = 18 minutes ). And of course k, the rate constant = 1.6

1 / [X] = ( 1.6 )( 18 minutes ) + 1 / ( 0.467 ) - Now let's solve for X

1 / [X] = 28.8 + 1 / ( 0.467 ),

1 / [X] = 28.8 + 2.1413...,

1 / [X] = 31,

[X] = 1 / 31 = ( About ) 0.03232 M

Now for this last bit here you probably are wondering why 1 / 31 is not 0.03232, rather 0.032258... Well, I did approximate one of the numbers along the way ( 2.1413... ) and took the precise value into account on my own and solved a bit more accurately. So that is your solution! The concentration of X after 18 minutes is about 0.03232 M

3 0

2 years ago