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2 years ago

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According to the kinetic theory, all matter is made of moving particles, which measurement of matter is directly proportional to

the
average kinetic energy of the particles?

1 answer:

34kurt2 years ago

8 0

Answer: Kelvin temperature of a substance

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Ipatiy [6.2K]

Your answer should be “C.”

MARK ME BRAINLIEST PLEASE!!!!!!

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3 years ago

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Pyridoxine contains 11 hydrogen atoms, 8 carbon atoms, 3 oxygen atoms, and 1 nitrogen atom. what is the chemical formula for pyr

FinnZ [79.3K]

Pyridoxine contains 11 hydrogen atoms, 8 carbon atoms, 3 oxygen atoms, and 1 nitrogen atom. The chemical formula would be C8H11NO3. A chemical formula of a substance tells the number of atoms of the elements that is involved in the compound. Pyrodoxine is commonly known as vitamin B6. It is mostly found in food like meat, whole grains, nuts and avocados. Also, it is can be present in dietary supplements. It has a number of important functions in the body. Dietary supplements are manufactured to prevent vitamin B6 deficiency. Also, it is used in treating a specific type of anemia or the lack of red blood cells.

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3 years ago

Which of the following examples of erosion would occur most quickly?

anyanavicka [17]

Answer: OC

Explanation:

The Others Could Take Many Millions Of Years

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3 years ago

What is the pH of pure water at 40.0Â°C if the Kw at this temperature is 2.92 - 10-147 OA) 6.767 O B) 0 465 O c) 7.000 OD) 7 233

Sergeeva-Olga [200]

Answer:

PH= 6.767 (answer is the A option)

Explanation:

first we need to correct the value in Kw at this temperature is 2.92*10^-14

so, in this case we have that:

Kw=2.92*10^-14 M²

[ H3O^+] [ H3O^+]

$[H_{3}O^{+} ] [OH^{-} ] = Kw = 2.92*10^{-14} M^{2} \\\\$

at 40ºC

$[H_{3}O^{+} ] = [OH^{-} ]$

$[H_{3}O^{+} ]^{2} = 2.92*10^{-14} M^{2}$

$[H_{3}O^{+} ] = (2.92*10^{-14})^{1/2} = 1.71*10^{-7} M$

$PH= -log10[H_{3}O^{+} ] = -log10(1.71*10^{-7} ) = 6.767$

7 0

2 years ago

Calculate E ° for the half‑reaction, AgCl ( s ) + e − − ⇀ ↽ − Ag ( s ) + Cl − ( aq ) given that the solubility product constant

antoniya [11.8K]

Answer: The value of $E^{o}$ for the half-cell reaction is 0.222 V.

Explanation:

Equation for solubility equilibrium is as follows.

$AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)$

Its solubility product will be as follows.

$K_{sp} = [Ag^{+}][Cl^{-}]$

Cell reaction for this equation is as follows.

$Ag(s)| AgCl(s)|Cl^{-}(0.1 M)|| Ag^{+}(1.0 M)| Ag(s)$

Reduction half-reaction: $Ag^{+} + 1e^{-} \rightarrow Ag(s)$ , $E^{o}_{Ag^{+}/Ag} = 0.799 V$

Oxidation half-reaction: $Ag(s) + Cl^{-}(aq) \rightarrow AgCl(s) + 1e^{-}$ , $E^{o}_{AgCl/Ag}$ = ?

Cell reaction: $Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)$

So, for this cell reaction the number of moles of electrons transferred are n = 1.

Solubility product, $K_{sp} = [Ag^{+}][Cl^{-}]$

= $1.77 \times 10^{-10}$

Therefore, according to the Nernst equation

$E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

At equilibrium, $E_{cell}$ = 0.00 V

Putting the given values into the above formula as follows.

$E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

$0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}$

$E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}$

= $0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}$

= 0.577 V

Hence, we will calculate the standard cell potential as follows.

$E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}$

$0.577 V = E^{o}_{Ag^{+}/Ag} - E^{o}_{AgCl/Ag}$

$0.577 V = 0.799 V - E^{o}_{AgCl/Ag}$

$E^{o}_{AgCl/Ag}$ = 0.222 V

Thus, we can conclude that value of $E^{o}$ for the half-cell reaction is 0.222 V.

3 0

2 years ago