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____ [38]
2 years ago
6

Shani poses as the Statue of Liberty for an art class. She holds a torch above her head with her right hand and a book against h

er side with her left hand. She holds the pose for ten minutes. At the end of the session, she mentions to her friend that it was a lot of work to pose and that her arms are really tired.
Which describes the work Shani did while she posed?(assume the starting work position is when her hand is already raised )

She did more work with her right hand than with her left because the right was held up higher.
She did no work while she posed because she did not move the torch or the book.
She did work while she posed because she was tired at the end of the session.
She did no work while she posed because she did not exert a force on anythi
Chemistry
2 answers:
viktelen [127]2 years ago
8 0
The first one, I believe!!! It is a fact that it took more energy for her left arm than her right arm to pose :) xxxx
uysha [10]2 years ago
6 0

Answer:

She did no work while she posed because she did not move the torch or the book.

Explanation:

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Given these reactions, where X represents a generic metal or metalloid 1) H2(g)+12O2(g)⟶H2O(g)ΔH1=−241.8 kJ 1) H2(g)+12O2(g)⟶H2O
Bond [772]

Answer:

ΔH = -793,6 kJ

Explanation:

It is possible to obtain ΔH of this reaction using Hess's law that says you can sum the half-reactions ΔH to obtain the ΔH of the global reaction:

If half-reactions are:

1) H₂(g) + ¹/₂O₂(g) ⟶ H₂O(g) ΔH₁ = −241.8 kJ

2) X(s) + 2Cl₂(g) ⟶ XCl₄(s) ΔH₂ = +356.9 kJ  

3) ¹/₂H₂(g) + ¹/₂Cl₂(g) ⟶ HCl(g) ΔH₃ = −92.3 kJ

4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄ = −639.1 kJ

5) H₂O(g) ⟶ H₂O(l) ΔH₅ = −44.0 kJ

The sum of (4) + 4×(3) - (2) - 2×(1) - 2×(5) is:

(4) X(s) + O₂(g) ⟶ XO₂(s) ΔH = −639.1 kJ

+4×(3) 2H₂(g) + 2Cl₂(g) ⟶ 4HCl(g) ΔH = −369,2 kJ

-(2) XCl₄(s) ⟶ X(s) + 2Cl₂(g) ΔH = -356,9 kJ

-2×(1) 2H₂O(g) ⟶ 2H₂(g) + O₂(g) ΔH = +483,6 kJ

-2×(5) 2H₂O(l) ⟶ 2H₂O(g) ΔH = +88.0 kJ

= <em>XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g)</em>

Where ΔH is:

ΔH = -639,1 kJ -369,2 kJ -356,9 kJ +483,6 kJ +88,0 kJ

<em>ΔH = -793,6 kJ</em>

I hope it helps!

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