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2 years ago

Why doesn’t the temperature of an object constantly rise as you add more and more kinetic energy?

Chemistry

1 answer:

Viefleur [7K]2 years ago

7 0

Because it is a non living things which means it doesn’t have and if the 5 senses we do

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Which element from this portion of the table chemically reacts in a way similar to the way the element chlorine (Cl) reacts? (1)

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D. I or Iodine

This is because they are both in the same periodic family (the halogens) and thus the number of valence electrons are the same

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3 years ago

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The website of a popular stargazers club notes that on Wednesday the moon will rise at 6:23 PM and set at 6:12 PM on Thursday. W

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how many moles of potassium hydroxide are needed to completely react with 1.73 miles of aluminum sulfate according to the follow

Masteriza [31]

<h3><u>Answer; </u></h3>

=10.38 moles KOH

<h3><u>Explanation</u>;</h3>

The balanced equation.

6KOH + Al2(SO4)3 --> 3K2SO4 + 2Al(OH)3

From the equation;

1 mole of aluminum sulfate requires 6 moles of potassium hydroxide.

Moles of Aluminium sulfate; 1.73 moles

Moles of KOH;

1 mol Al2(SO4)3 : 6 mol KOH = 1.73 mol Al2(SO4)3 : x mol KOH

Thus; x = (6 × 1.73)

<u> =10.38 moles KOH </u>

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3 years ago

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Which half-reaction is most easily oxidized?

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3 years ago

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10.0 g Cu, C Cu = 0.385 J/g°C 10.0 g Al, C Al = 0.903 J/g°C 10.0 g ethanol, Methanol = 2.42 J/g°C 10.0 g H2O, CH2O = 4.18 J/g°C

Mazyrski [523]

Answer:

Lead shows the greatest temperature change upon absorbing 100.0 J of heat.

Explanation:

$Q=mc\Delte T$

Q = Energy gained or lost by the substance

m = mass of the substance

c = specific heat of the substance

ΔT = change in temperature

1) 10.0 g of copper

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the copper = c = 0.385 J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 0.385J/g^oC}=25.97^oC$

2) 10.0 g of aluminium

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the aluminium= c = 0.903 J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 0.903 J/g^oC}=11.07^oC$

3) 10.0 g of ethanol

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the ethanol= c = 2.42 J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 2.42 J/g^oC}=4.13 ^oC$

4) 10.0 g of water

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the water = c = 4.18J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 4.18 J/g^oC}=2.39 ^oC$

5) 10.0 g of lead

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the lead= c = 0.128 J/g°C

$\Delta T=\frac{Q}{mc}$

$=\frac{100.0 J}{10 g\times 0.128 J/g^oC}=78.125^oC$

Lead shows the greatest temperature change upon absorbing 100.0 J of heat.

3 0

3 years ago