Answer:
Half-reactions:
Cr³⁺ + 1e⁻ → Cr²⁺; Zn → Zn²⁺ + 2e⁻
Net ionic equation:
2Cr³⁺ + Zn → 2Cr²⁺ + Zn²⁺
Explanation:
The Cr³⁺ is reduced to Cr²⁺:
<h3>
Cr³⁺ + 1e⁻ → Cr²⁺ -Half-reaction 1-</h3>
Zn is oxidized to Zn²⁺:
<h3>
Zn → Zn²⁺ + 2e⁻ -Half-reaction 2-</h3>
Twice the reduction of Cr:
2Cr³⁺ + 2e⁻ → 2Cr²⁺
Now this reaction + Oxidation of Zn:
2Cr³⁺ + 2e⁻ + Zn → 2Cr²⁺ + Zn²⁺ + 2e⁻
<h3>2Cr³⁺ + Zn → 2Cr²⁺ + Zn²⁺ - Net ionic equation</h3>
Answer:
0.252 mol
Explanation:
<em>Given the following reaction: </em>
<em>Cu + 2 AgNO₃ → 2 Ag + Cu(NO₃)₂</em>
<em>How many moles of Ag will be produced from 16.0 g Cu, assuming AgNO3 is available in excess.</em>
First, we write the balanced equation.
Cu + 2 AgNO₃ → 2 Ag + Cu(NO₃)₂
We can establish the following relations.
- The molar mass of Cu is 63.55 g/mol.
- The molar ratio of Cu to Ag is 1:1.
The moles of Ag produced from 16.0 g of Cu are:

NaCl and H2S will experience dipole-dipole interaction because they are permanently polarized.
Answer:
A) photon
Explanation:
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