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3 years ago

I’m trying to figure out how to convert 6 moles KCL to particles.

Chemistry

1 answer:

I am Lyosha [343]3 years ago

7 0

Answer:

How many moles KCl in 1 grams? The answer is 0.013413582325

1 mole is equal to 1 moles KCl, or 74.5513 grams.

447.3078 is the answer

Explanation:

~Cornasha_Weeb

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Calculate the ph of sweater which has a hydrogen ion concentration of 1 times 10 to the power of 8

telo118 [61]

Answer:

The pH of the sweater containing Hydrogen ion concentration

$[H^{+}]=1\times 10^{-8}$ is

8

Explanation:

pH = It is the negative logarithm of activity (concentration) of hydrogen ions.

pH = -log([H+])

Now, In the question the concentration of [H+] ions is :

$[H^{+}]=1\times 10^{-8}$

$pH = -log(1\times 10^{-8})$

use the relation:

$log(10^{-a})=a$

$pH= -(-8)$

pH = 8

Note : 1 times 10 to the power of 8 must be" 1 times 10 to the power of -8"

If the concentration is

$[H^{+}]=1\times 10^{8}$

Then pH = -8 , which is not possible . So in that case the pH calculation is by other method

5 0

3 years ago

A rod shaped organism with no nucleus commonly found in Garbage will most likely belong to which kingdom

Dovator [93]

Answer:

Monera

Explanation:

This organism is only bacteria (bacilli bacteria). They are simple unicellular organism; they have no nucleus. They barely have organelles. These organism in the garbage are heterotrophic /saprophytic and cause the composition of the organic molecules (such as the discarded foods).

8 0

3 years ago

Determine the free energy(ΔG) from the standard cell potential (Ecell0 ) for the reaction:2ClO2-(aq)+Cl2(g)→2ClO2(g)+ 2Cl-(aq)wh

Dima020 [189]

Answer: The $\Delta G^o$ for the given reaction is $-7.84\times 10^4J$

Explanation:

For the given chemical reaction:

$2ClO_2^-(aq.)+Cl_2(g)\rightarrow 2ClO_2(g)+2Cl^-(aq.)$

Half reactions for the given cell follows:

Oxidation half reaction: $ClO_2^-\rightarrow ClO_2+e^-;E^o_{ClO_2^-/ClO_2}=0.954V$ ( × 2)

Reduction half reaction: $Cl_2+2e^-\rightarrow 2Cl(g);E^o_{Cl_2/2Cl^-}=1.36V$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=1.36-(0.954)=0.406V$

To calculate standard Gibbs free energy, we use the equation:

$\Delta G^o=-nFE^o_{cell}$

Where,

n = number of electrons transferred = 2

F = Faradays constant = 96500 C

$E^o_{cell}$ = standard cell potential = 0.406 V

Putting values in above equation, we get:

$\Delta G^o=-2\times 96500\times 0.406=-78358J=-7.84\times 10^4J$

Hence, the $\Delta G^o$ for the given reaction is $-7.84\times 10^4J$

4 0

3 years ago

If the energy of photon emitted from the hydrogen atom is 4.09 x 10-19 J, what is

Aliun [14]

Answer:

486 nm

Explanation:

From the question given above, the following data were obtained:

Energy (E) = 4.09×10¯¹⁹ J

Wavelength (λ) =?

Next, we shall determine the frequency of the photon. This can be obtained as follow:

Energy (E) = 4.09×10¯¹⁹ J

Planck's constant (h) = 6.63×10¯³⁴ Js

Frequency (f) =?

E = hf

4.09×10¯¹⁹ = 6.63×10¯³⁴ × f

Divide both side by 6.63×10¯³⁴

f = 4.09×10¯¹⁹ / 6.63×10¯³⁴

f = 6.17×10¹⁴ Hz

Next, we shall determine the wavelength of the photon. This can be obtained as follow:

Frequency (f) = 6.17×10¹⁴ Hz

Velocity of photon (v) = 3×10⁸ m/s

Wavelength (λ) =?

v = λf

3×10⁸ = λ × 6.17×10¹⁴

Divide both side by 6.17×10¹⁴

λ = 3×10⁸ / 6.17×10¹⁴

λ = 4.86×10¯⁷ m

Finally, we shall convert 4.86×10¯⁷ m to nm. This can be obtained as follow:

1 m = 1×10⁹ nm

Therefore,

4.86×10¯⁷ m = 4.86×10¯⁷ m × 1×10⁹ nm / 1 m

4.86×10¯⁷ m = 486 nm

Therefore, the wavelength of the photon is 486 nm

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3 years ago

A particular power source creates electricity through the use of giant magnets that are rotated to create a flow of electrons. T

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From the electric generator to electric outlets in homes

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3 years ago

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