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kondaur [170]
3 years ago
12

What is an orbital shell

Chemistry
1 answer:
Lyrx [107]3 years ago
4 0
<span>orbital shell is the</span><span> the circular paths around the nucleus of an atom along which the electrons traverse.</span>
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C would be the right answer

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One theory that has been proposed about the information of the universe.
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The big bang theory is one of the multipl about the universe

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At 60°c, the ion-product constant of water, kw, is 9.25 × 10–14 . what is the ph of pure water at 60°c ?
Rus_ich [418]

Answer:

6.517.

Explanation:

  • It is known that for pure water: [H₃O⁺] = [OH⁻], so water is neutral.  

  • It also is known that:

<em>Kw = [H₃O⁺][OH⁻] = 9.25 x 10⁻¹⁴.</em>

∵ [H₃O⁺] = [OH⁻].

∴ [H₃O⁺]² = 9.25 x 10⁻¹⁴.

∴ [H₃O⁺] = √(9.25 x 10⁻¹⁴) = 3.041 x 10⁻⁷ M.

∵ pH = - log[H₃O⁺]

<em>∴ pH</em> = - log(3.041 x 10⁻⁷) = <em>6.517.</em>

3 0
3 years ago
Explain whether each of the possible reactions listed below will occur. For those that will, predict what the products will be.
Alex777 [14]

Answer:

Zn(s) + 2H₂O (l) → Zn(OH)2(aq) + H2(g)

Sn(s) + O2(g) →SnO2(s)

Cd(s) + Pb(NO3)2 (aq) →Cd(NO3)2 (aq) + Pb(s)

Cu (s) + HCl(aq) → can not occur

Explanation:

Metals reacts with liquid water to yield the corresponding metallic hydroxide and hydrogen gas as shown above; Zn(s) + 2H₂O (l) → Zn(OH)2(aq) + H2(g)

Tin reacts with oxygen to yield tin (IV) oxide. This is an oxidation reaction as shown; Sn(s) + O2(g) →SnO2(s)

The reaction between Cd and Pb(NO3)2 is a single replacement reaction. It is possible because Cd is above Pb in the electrochemical series. Hence the reaction occurs thus; Cd(s) + Pb(NO3)2 (aq) →Cd(NO3)2 (aq) + Pb(s)

Copper does not displace hydrogen from dilute acids (such as HCl) since copper is lower than hydrogen in the electrochemical series hence the reaction does  not occur.

3 0
3 years ago
Determine the molecular mass ratio of two gases whose rates of effusion have a ratio of 16 : 1.
Usimov [2.4K]
Answer is: <span>the molecular mass ratio of two gases is 1 : 256.
</span>rate of effusion of gas1 : rate of effusion of gas = 16 : 1.<span>
rate of effusion of gas1 = 1/√M(gas1).
rate of effusion of gas2 = 1/√M(gas2).
rate of effusion of gas1 = rate of effusion of gas2 </span>· 16<span>.
</span>1/√M(gas1) = 1/√M(gas2) · 16 /².
<span>1/M(gas1) = 1/M(gas2) </span>· 256.
<span>M(gas1) </span>· 256 = M(gas2).<span>

</span>
8 0
3 years ago
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