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babunello [35]
3 years ago
11

Which event always involves a chemical change A boiling B melting C conducting D burning

Chemistry
1 answer:
Andrew [12]3 years ago
4 0
The answer is Burning
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If I have a 200 L container filled with nitrogen at a pressure of 1.0 atm, how many moles
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3 years ago
What energy is required to remove the remaining electron from singly ionized helium?
skelet666 [1.2K]

To remove one electron from singly ionized helium, will require approximately 54.4 eV or 8.72 1020 J of energy.

The amount of energy required by an isolated, gaseous molecule in the electronic state of the ground to absorb in order to discharge an electron and produce a cation has been known as the ionization energy. The amount of energy required for every atom in a mole to drop one electron is most often given as kJ/mol.

Anything that causes electrically neutral atoms and molecules to gain or lose electrons in order to become electrically charged atoms as well as molecules .

Therefore, the "To remove one electron from singly ionized helium, will require approximately 54.4 eV or 8.72 1020 J of energy."

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3 0
1 year ago
How many of these transition metals have you heard of before?
AleksAgata [21]

Answer:

38 elements or Element 31: Gallium Element 32: Germanium Element 33: Arsenic Element 34: Selenium Element 35: Bromine Element .

Explanation:

How many transition metals are there?

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6 0
2 years ago
A lead mass is heated and placed in a foam cup calorimeter containing 40.0 mL of water at 17.0°C. The water reaches a temperatur
lbvjy [14]

Answer: 502 Joules

Explanation:

To calculate the mass of water, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of water = 1 g/mL

Volume of water = 40.0 mL

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{40.0mL}\\\\\text{Mass of water}=(1g/mL\times 40.0mL)=40.0g

When metal is dipped in water, the amount of heat released by lead will be equal to the amount of heat absorbed by water.

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

q=m\times c\times \Delta T

q = heat absorbed by water

m = mass of water = 40.0 g

T_{final} = final temperature of water = 20.0°C

T_{initial = initial temperature of water = 17.0°C

c = specific heat of water= 4.186 J/g°C

Putting values in equation 1, we get:

q=40.0\times 4.186\times (20.0-17.0)]

q=502J

Hence, the joules of heat were re-leased by the lead is 502

5 0
3 years ago
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