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Molodets [167]
3 years ago
10

Someone help asap its due in 10 mins

Chemistry
1 answer:
solniwko [45]3 years ago
4 0

Answer:

A. Species

Explanation:

The order from least to most specific is:

Kingdom

Phylum

Class

Order

Family

Genus

Species

You might be interested in
How many miles of lead are equal to 9.51 x 10^3 g Pb?
Lemur [1.5K]
Molar mass Pb = 207.2 g/mol

1 mole Pb ------------- 207.2
? mole Pb ------------ 9.51 x 10³

moles = 9.51 x 10³ * 1 / 207.2

moles = 9.51 x 10³ / 207.2

= 45.89 moles

hope this helps!

7 0
3 years ago
Explain in detail what causes tides and give one reason why it is important for humans to monitor tides.
abruzzese [7]
If the moon is on the other side of the earth, the water is getting pulled back because of the gravity force but if the moon is on the same side of the beach it's high tide.
8 0
3 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
3 years ago
Changing the number of<br> element<br> produces isotopes of the same
Ber [7]

Answer:

An isotope is one of two or more forms of the same chemical element. Different isotopes of an element have the same number of protons in the nucleus, giving them the same atomic number, but a different number of neutrons giving each elemental isotope a different atomic weight.

3 0
2 years ago
What happens to particles during<br> changes of state between solids,<br> liquids, and gases?
oksano4ka [1.4K]
Answer:
Their vibrations speed up

Explanation:
They start vibrating faster and faster and start generating more and more heat and separate from each other so, therefore (usually), become less dense
7 0
3 years ago
Read 2 more answers
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