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Masteriza [31]
3 years ago
7

2. How much energy must be removed from a 50.0 g sample of steam in order to get it to condense

Chemistry
1 answer:
Strike441 [17]3 years ago
5 0

Answer:

113 kJ or 113,000J

Explanation:

Since we are condensing it, a physical change in the state of matter is occurring. 113 kJ or 113,000J because of the following equation: Q = 2260 times 50.  2260 is the Heat of Vaporization constant.

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Because all matter has density and since air has density then air is matter and plus EVERYTHING is made of matter
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How many grams Kl are needed to prepare 100.mL of 0.55M solution?​
Makovka662 [10]

Answer:

Explanation:

100mL = 0.1L

0.55 M = mol/0.1 L

mol = 0.055 mol

molar mass of KI = 165.998 g

0.055 * 165.998 = 9.13 g of KI

7 0
2 years ago
PLEASE HELP!!! WILL MARK BRAINLIEST!
fredd [130]

The longest chain of carbons will contain the highest number of carbon atoms.

Explanation:

question 3 - the longest chain have 7 carbon atoms

question 4 - the longest chain have 8 carbon atoms

question 1 - the longest chain have 5 carbon atoms

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longest carbon chain

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5 0
3 years ago
Calculate the energy that is required to change 50.0 g ice at -30.0°C to a liquid at 73.0°C. The heat of fusion = 333 J/g, the h
OverLord2011 [107]

Answer:

There is 3.5*10^4 J of energy needed.

Explanation:

<u>Step 1:</u> Data given

Mass of ice at -30.0 °C = 50.0 grams

Final temperature = 73.0 °C

The heat of fusion = 333 J/g

the heat of vaporization = 2256 J/g

the specific heat capacity of ice = 2.06 J/gK

the specific heat capacity of liquid water = 4.184 J/gK

<u>Step 2:</u> Calculate the heat absorbed by ice

q = m*c*(T2-T1)

⇒ m = the mass of ice = 50.0 grams

⇒ c = the heat capacity of ice = 2.06 J/gK = 2.06 J/g°C

⇒ T2 = the fina ltemperature of ice = 0°C

⇒ T1 = the initial temperature of ice = -30.0°C

q = 50.0 * 2.06 J/g°C * 30 °C

q = 3090 J

<u>Step 3:</u> Calculate heat required to melt the ice at 0°C:

q = m*(heat of fusion)

q = 50.0* 333J/g

q =  16650 J

<u> </u>

<u>Step 4</u>: Calculate the heat required to raise the temperature of water from 0°C to 73.0°C

q = m*c*(T2-T1)

 ⇒ mass = 50.0 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = T2-T1 = 73.0 - 0  = 73 °C

q = 50.0 * 4.184 * 73.0 = 15271.6 J

<u>Step 5:</u> Calculate the total energy

qtotal = 3090 + 16650 + 15271.6 = 35011.6 J = 3.5 * 10^4 J

There is 3.5*10^4 J of energy needed.

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3 years ago
What organ synthesizes the largest quantity of cholesterol?
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