3 years ago

21.

Chemistry

1 answer:

Bond [772]3 years ago

7 0

The answer is A. The pH decreases

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What is the change in energy of an atom if the wavelength of the photon absorbed by the atom is equal to 4.64 x 10-7 meters? (Pl

Dmitry [639]

Answer : The change in energy of an atom is, $4.28\times 10^{-19}J$

Solution : Given,

Wavelength of the photon = $4.64\times 10^{-7}m$

Formula used :

$E=h\times \frac{c}{\lambda}$

where,

E = energy of an atom

h = Planck’s constant = $6.626\times 10^{-34}Js$

c = speed of light = $2.998\times 10^{8}m/s$

$\lambda$ = wavelength of photon

Now put all the given values in the above formula, we get the energy of an atom.

$E=(6.626\times 10^{-34}Js)\times \frac{2.998\times 10^8m/s}{4.64\times 10^{-7}m}$

$E=4.28\times 10^{-19}J$

Therefore, the change in energy of an atom is, $4.28\times 10^{-19}J$

7 0

4 years ago

Read 2 more answers

Given the standard entropy values in the table, what is the value of º for the following reaction?

Mandarinka [93]

Hello.

The answer is <span>+313.766 J/mol·K
</span>
Use the coefficients of the reaction and sum the product entropies less the reactant entropies:

4*188.8 + 2*213.7 - 3*205.1 - 2* 126.8 = 313.7 J/mol*K

Have a nice day

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4 years ago

Deceive the photoelectric effect and explain why it made modifications to the Rutherford model necessary

Anna [14]

Nooooooooooooooooooo

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3 years ago

Calculate the molarity of a solution that contains 4.7 moles in 5.2 liters.

AleksAgata [21]

Answer:

0.9M

Explanation:

molarity = mols/L

M=4.7mol/5.2L

= 0.9M

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3 years ago

Enter your answer in the provided box. The vapor pressure of ethanol is 1.00 x 10² mmHg at 34.90°C. What is its vapor pressure a

zavuch27 [327]

Answer:

2,54x10² mmHg

Explanation:

To solve this problem you can use Clausius-Clapeyron equation that serves to estimate vapor pressures or temperatures:

$Ln(\frac{P_{2}}{P_{1}} ) =\frac{ deltaH_{vap}}{R} (\frac{1}{T_{1}}-\frac{1}{T_{2}} )$

Where:

P1 is 1,00x10² mmHg

ΔHvap is 39,3 kJ/mol

R is gas constant 8,314x10⁻³ kJmol⁻¹K⁻¹

T1 is 34,90°C + 273,15 = 308,05 K

T2 is 54,81°C + 273,15 = 327,96 K

Thus:

$Ln(\frac{P_{2}}{1,0x10^{2}mmHg}) =\frac{39,3kJ/mol}{8,314x10^{-3}kJ/molK} (\frac{1}{308,05K}-\frac{1}{327,96K} )$

Thus, P2 is <em>2,54x10² mmHg</em>

I hope it helps!

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4 years ago