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Dahasolnce [82]
3 years ago
11

Change 23 degrees Celsius to F

Chemistry
2 answers:
Fed [463]3 years ago
8 0
Rewrite the formula C=5/9(F-32) substituting 23 for C: 23=5/9(F-32), then multiply both sides by the reciprocal of 5/9.
(9/5)*(23)=(9/5)*5/9(F-32)
41.4=F-32; add 32 to both sides.
41.4+32=F-32+32
73.4=F
Marat540 [252]3 years ago
7 0
It's 73.5 F cz c/5=f-32/9
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The change in entropy is related to the change in the number of moles of gas molecules. Determine the change in moles of gas for
tester [92]

The given question is incomplete. The complete question is:

The change in entropy is related to the change in the number of moles of gas molecules. Determine the change in moles of gas for each of the reactions and decide if the entropy increases decreases or has little to no change:

A. K(s)+O_2(g)\rightarrow KO_2(s)

B. CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g)

C. CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)

D. N_2O_2(g)\rightarrow 2NO(g)+O_2(g)

Answer: A. K(s)+O_2(g)\rightarrow KO_2(s) : decreases

B. CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2(g) : decreases

C. CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g): no change

D. N_2O_2(g)\rightarrow 2NO(g)+O_2(g) : increases

Explanation:

Entropy is defined as the randomness of the system.

Entropy is said to increase when the randomness of the system increase, is said to decrease when the randomness of the system decrease and is said to have no change when the randomness remains same.

In reaction K(s)+O_2(g)\rightarrow KO_2(s), as gaseous reactant is changed to solid product, entropy decreases.

In reaction CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g), as 4 moles of gaseous reactants is changed to 2 moles of gaseous product, entropy decreases.

In reaction CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g), as 3 moles of gaseous reactants is changed to 3 moles of gaseous product, entropy has no change.

In reaction  N_2O_2(g)\rightarrow 2NO(g)+O_2(g) , as 1 mole of gaseous reactant is changed to 3 moles of gaseous product, entropy increases.

7 0
3 years ago
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8090 [49]
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After identifying each states, let's investigate the energy for phase change. Let's start with the arrows pointing to the right. The first arrow to the right is a phase change from solid to liquid. The intermolecular forces in a solid is the strongest among the three phases of matter. So, you would need an input of energy to break them apart into liquid. The same is true for the phase change from liquid to gas. Therefore, all the arrows pointing to the right require an input of energy.

The reverse arrows pointing to the left needs to release energy. The molecules in the gas state are free such that they can travel from one point to another easily. They have the highest amount of energy. So, if you want the molecules to come closer together, you need to remove the energy to keep them in place. Therefore, the arrows pointing to the right require removal of energy.
7 0
3 years ago
Read 2 more answers
Chlorophyll a is one of the green pigments found in plants. Chlorophyll a has the molecular formula C55H72MgN4O5. How many atoms
algol13
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8 0
2 years ago
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tatuchka [14]

Explanation:

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6 0
3 years ago
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OLga [1]
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This statement is false. A CH4 molecule do not have a hydrogen bonding instead it has dipole dipole attraction.

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This would be a true statement. A hydrogen bond is present when an atom of hydrogen shares electrons with O, N or F atom.

A hydrogen bond is equivalent to a covalent bond.
This is a false statement. A hydrogen bond is an intermolecular force of attraction while covalent bond is a intramolecular force. So, they would mean different things.

a hydrogen bond is possible with only certain hydrogen-containing compounds.
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a hydrogen atom acquires a partial positive charge when it is covalently bonded to an f atom.
This would be true since a HF is a polar molecule.
6 0
3 years ago
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