CHEMISTRY IN FOCUS (LL)-TEXT
CHEMISTRY IN FOCUS (LL)-TEXT
7th Edition
ISBN: 9781337399845
Author: Tro
Publisher: CENGAGE L
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Chapter 10, Problem 32E
Interpretation Introduction

Interpretation:

The average solar power in kilowatts and the number of kilowatt-hours that fall on the rooftop in one year are to be calculated and whether the roof receives enough energy to meet the energy needs or not and the minimum percent efficiency of the solar cells are to be determined.

Concept Introduction:

Power input on a PV cell is the total active area of the PV cell multiplied by the solar power.

The percent of the input power that is converted into the output power by a PV cell is called its efficiency.

Percent efficiency of a PV cell is 100 multiplied by the ratio of power out and power in.

% efficiency=power outpower in×100% …… (1)

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AG/F-2° V 3. Before proceeding with this problem you may want to glance at p. 466 of your textbook where various oxo-phosphorus derivatives and their oxidation states are summarized. Shown below are Latimer diagrams for phosphorus at pH values at 0 and 14: -0.93 +0.38 -0.50 -0.51 -0.06 H3PO4 →H4P206 →H3PO3 →→H3PO₂ → P → PH3 Acidic solution Basic solution -0.28 -0.50 3--1.12 -1.57 -2.05 -0.89 PO HPO H₂PO₂ →P → PH3 -1.73 a) Under acidic conditions, H3PO4 can be reduced into H3PO3 directly (-0.28V), or via the formation and reduction of H4P206 (-0.93/+0.38V). Calculate the values of AG's for both processes; comment. (3 points) 0.5 PH P 0.0 -0.5 -1.0- -1.5- -2.0 H.PO, -2.3+ -3 -2 -1 1 2 3 2 H,PO, b) Frost diagram for phosphorus under acidic conditions is shown. Identify possible disproportionation and comproportionation processes; write out chemical equations describing them. (2 points) H,PO 4 S Oxidation stale, N
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