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Vilka [71]
3 years ago
14

Help please ill do anything really!!!!!!!

Chemistry
1 answer:
Volgvan3 years ago
5 0

Answer:

The first one is air the second is decreases the third is water the fourth is gas and the last is liquid.

Explanation:

Hope it helps.

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Describe in detail what you know about the enthalpy, entropy, and free energy changes when a sample of gas condenses to a liquid
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The keyword here is gas condenses to a liquid, which mean we're talking about condensation process

The enthalpy energy in condensation process is negative because it releases energy
The entropy in general will also decreases .

Temperature affect this change because it will create free energy if added with this result

hope this helps
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2. Which test for iron(II) ions is conclusive ​
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Answer:

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Explanation:

The addition of K 3 Fe(CN) 6 to a solution causes the formation of a deep blue precipitate which indicates that iron(II) ions are present.

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How many valence electrons do nitrogen atoms have
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Question 6 (1 point)
Mnenie [13.5K]

Answer:

The correct option is;

d 4400

Explanation:

The given parameters are;

The mass of the ice = 55 g

The Heat of Fusion = 80 cal/g

The Heat of Vaporization = 540 cal/g

The specific heat capacity of water = 1 cal/g

The heat required to melt a given mass of ice = The Heat of Fusion × The mass of the ice

The heat required to melt the 55 g mass of ice = 540 cal/g × 55 g = 29700 cal

The heat required to raise the temperature of a given mass ice (water) = The mass of the ice (water) × The specific heat capacity of the ice (water) × The temperature change

The heat required to raise the temperature of the ice from 0°C to 100°C = 55 × 1 × (100 - 0) = 5,500 cal

The heat required to vaporize a given mass of ice = The Heat of Vaporization × The mass of the ice

The heat required to vaporize the 55 g mass of ice at 100°C = 80 cal/g × 55 g = 4,400 cal

The total heat required to boil 55 g of ice = 29700 cal + 5,500 cal + 4,400 cal = 39,600 cal

However, we note that the heat required to vaporize the 55 g mass of ice at 100°C = 80 cal/g × 55 g = 4,400 cal.

The heat required to vaporize the 55 g mass of ice at 100°C = 4,400 cal

3 0
3 years ago
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