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LUCKY_DIMON [66]
3 years ago
12

A solution has a higher boiling point than its associated pure solvent does.

Chemistry
1 answer:
marta [7]3 years ago
7 0

Answer:

4 boiling point elevation

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The reaction of hemoglobin with oxygen can be written as follows. hb + o2 â hbo2 if the amount of oxygen available to the blood
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Oxygen is needed to carry out a lot of biochemical processes in the body. If the amount of oxygen available to the blood decreases significantly a lot of things will go wrong in the body. For instance, lack of adequate oxygen will lead to the death of neurons which will eventually leads to brain cells death and irreparable brain damage. Oxygen is also needed for cellular respiration, without respiration, there will not be oxygen for carrying out various cellular activities and this will result into death. Oxygen deprivation will also leads to difficulty in breathing and other associated problems.
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How many valence electrons does an atom of Iodine have?<br><br> a) 5 <br> b) 6 <br> c) 7 <br> d) 8
baherus [9]
The answer is 7 !! :)
8 0
2 years ago
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C10H22 + O2 -- CO2 + H2O balance
Tasya [4]

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1,31÷2 =10,11

Explanation:

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What is the relationship between mass and gravity
Tpy6a [65]

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since the gravitational force is directly proportional to the mass of both interacting objects, more massive objects will attract each other with a greater gravitational force. So as the mass of either object increases, the force of gravitational attraction between them also increases.

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5 0
2 years ago
Consider the formation of [Ni(en)3]2+ from [Ni(H2O)6]2+. The stepwise ΔG∘ values at 298 K are ΔG∘1 for first step=−42.9 kJ⋅mol−1
timurjin [86]

Answer:

kf = 1.16 x 10¹⁸

Explanation:

Step 1: [Ni(H₂O)₆]²⁺  + 1en → [Ni(H₂O)₄(en)]²⁺  ΔG°1 = -42.9 kJmol⁻¹

Step 2: [Ni(H₂O)₄(en)]²⁺  + 1en → [Ni(H₂O)₂(en)₂]²⁺  ΔG°2 = -35.8 kJmol⁻¹

Step 3: [Ni(H₂O)₂(en)₂]²⁺ + 1en →  [Ni(en)₃]²⁺  ΔG°3 = -24.3 kJmol⁻¹

________________________________________________________

Overall reaction: [Ni(H₂O)₆]²⁺  + 3en → [Ni(en)₃]²⁺  ΔG°r

ΔG°r = ΔG°1 + ΔG°2 + ΔG°3

ΔG°r = -42.9 - 35.8 - 24.3

ΔG°r = -103.0 kJmol⁻¹

ΔG°r = -RTlnKf

-103,000 Jmol⁻¹ =  - 8.31 J.K⁻¹mol⁻¹ x 298 K x lnKf

kf = e ^(-103,000/-8.31x298)

kf = e ^41.59

kf = 1.16 x 10¹⁸

7 0
3 years ago
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