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lukranit [14]
3 years ago
13

Would a burning candle have potential or kinetic energy?

Chemistry
1 answer:
Margarita [4]3 years ago
3 0

Answer:

When the candle is lit and burning, it has kinetic energy.

Explanation:

please smash brainliest on my doorstep

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The relationship between pressure and volume, when moles and temperature of a gas are held constant, is: PV = k. We could say th
pochemuha
You can say that if the volume of the gas is halved, the pressure is doubled.

The expression shows that pressure and volume are inversely proportional if temperature and amount of gas is held constant.  This means that if volume goes down the pressure needs to go up.  That also means that in order to maintain the K value, if pressure is doubled the volume needs to be halved and if the pressure is halved the volume needs to be doubled.

This relationship only works if we assume everything else (Temperature and moles of gas) to be constant.
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In a chemical formula the right side is called?
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HLP PLZ. ____ g iron reacts with 71 g chlorine to produce 129 g of iron (II) chloride.
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Answer: 58

Explanation:

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3 years ago
What does this look like? Can someone write out the Stoichiometry starting with 50 grams of carbon (use balanced equation)
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3 years ago
A chunk of tin weighing 18.5 grams and originally at 97.38 °C is dropped into an insulated cup containing 75.7 grams of water at
weqwewe [10]

Answer:

22.44°C will be the final temperature of the water.

Explanation:

Heat lost by tin will be equal to heat gained by the water

-Q_1=Q_2

Mass of tin = m_1=18.5 g

Specific heat capacity of tin = c_1=0.21 J/g^oC

Initial temperature of the tin = T_1=97.38^oC

Final temperature = T_2=T

Q_1=m_1c_1\times (T-T_1)

Mass of water= m_2=75.7 g

Specific heat capacity of water= c_2=4.184 J/g^oC

Initial temperature of the water = T_3=21.52^oC

Final temperature of water = T_2=T

Q_2=m_2c_2\times (T-T_3)

-Q_1=Q_2

-(m_1c_1\times (T-T_1))=m_2c_2\times (T-T_3)

On substituting all values:

-(18.5 g\times 0.21 J/g^oC\times (T-97.38^oC))=75.7 g\times 4.184 J/g^oC\times (T-21.52 ^oC)

we get, T = 22.44°C

22.44°C will be the final temperature of the water.

5 0
4 years ago
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