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Alla [95]
3 years ago
14

Which arrow represents the change of state described above? LNOP​

Chemistry
2 answers:
Thepotemich [5.8K]3 years ago
7 0

Answer:

P Is your winner uWu

Explanation:

Be happy :)

denis-greek [22]3 years ago
6 0

Answer:

The answer is either o or p i think it is p

Explanation:

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What mass of oxygen is needed to combust and produced 6.4 moles of carbon dioxide gas?
N76 [4]
You have already gotten the balanced equation. And the ratio of mol number of reactants and production is the ratio of coefficient. So there is 6.4/8*11=8.8 mol oxygen needed. The mass is 8.8*32=281.6 g.
6 0
3 years ago
2. Name each of the following ionic compounds A. K2O B. Cacl2. C. Mg3N2 D. NaCIO E. KNO3​
Ludmilka [50]
A. Potassium oxide
B. Calcium chloride
C. Magnesium nitride
D. Sodium hypochlorite
E. Potassium nitrate
3 0
3 years ago
PLSSS HELP MEE 100 POINTS!!!!!!! How many atoms of each element does magnesium hydroxide, Mg(OH)2, contain? Magnesium = 1 atom,
Basile [38]

It would be Magnesium = 1 atom, Oxygen = 2 atoms, Hydrogen = 2 atoms. It has 5 atoms total. There is 1 magnesium and you multiply each element by the the 2 outside the parenthesis to get 2 oxygen and 2 hydrogen.

5 0
3 years ago
Read 2 more answers
What is the molar mass of sulfur (S)?
Lesechka [4]

Answer:

A. 32.06 g/mol

Explanation:

The molar mass units are always g/mol

8 0
2 years ago
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Whats the voltage of CuCl2 + Zn -> ZnCl2 + Cu
gtnhenbr [62]

Answer:

Approximately 1.10\; {\rm V} under standard conditions.

Explanation:

Equation for the overall reaction:

{\rm CuCl_{2}}\, (aq) + {\rm Zn}\, (s) \to {\rm ZnCl_{2}} \, (aq) + {\rm Cu}\, (s).

Write down the ionic equation for this reaction:

\begin{aligned}& {\rm Cu^{2+}}\, (aq) + 2\; {\rm Cl^{-}}\, (aq) + {\rm Zn}\, (s)\\ & \to {\rm Zn^{2+}} \, (aq) + 2\; {\rm Cl^{-}}\, (aq) + {\rm Cu}\, (s)\end{aligned}.

The net ionic equation for this reaction would be:

{\rm Cu^{2+}}\, (aq) + {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + {\rm Cu}\, (s).

In this reaction:

  • Zinc loses electrons and was oxidized (at the anode): {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}}.
  • Copper gains electrons and was reduced (at the cathode): {\rm Cu^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Cu} \, (s).

Look up the standard potentials for each half-reaction on a table of standard reduction potentials.

Notice that {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} is oxidation and is likely not on the table of standard reduction potentials. However, the reverse reaction, {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s), is reduction and is likely on the table.

  • E(\text{anode}) = -0.7618\; {\rm V} for {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s), and
  • E(\text{cathode}) = 0.3419\; {\rm V} for {\rm Cu^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Cu} \, (s).

The reduction potential of {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} would be -E(\text{anode}) = -(-0.7618\; {\rm V}) = 0.7618\; {\rm V}, the opposite of the reverse reaction {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s).

The standard potential of the overall reaction would be the sum of the standard potentials of the two half-reactions:

\begin{aligned} E^{\circ} &= E^{\circ}(\text{cathode}) + (-E^{\circ}(\text{anode})) \\ &= 0.3419 - (-0.7618\; {\rm V}) \\ &\approx 1.10\; {\rm V}\end{aligned}.

7 0
2 years ago
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