Answer:

Explanation:
Given that:

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

Multiplying (2) with equation (4) ; we have:

From equation (1) ; multiplying (-1) with equation (1); we have:

From equation (2); multiplying (3) with equation (2); we have:

Now; Adding up equation (5), (6) & (7) ; we get:



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
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(According to Hess Law)


Answer: 5
Explanation: add up all the electrons and it will amount to 23. Arranging by the old model for electronic configuration, we have : 2, 8, 8, 5
The last number being 5 represent its valence electron
An ecosystem is a geographic area where plants, animals, and other organisms, as well as weather and landscape, work together to form a bubble of life. Ecosystems contain biotic or living, parts, as well as abiotic factors, or nonliving parts. Biotic factors include plants, animals, and other organism .
Answer:
<h2>Percentage error is a measurement of the discrepancy between an observed and a true, or accepted value .</h2>
Explanation:
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