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Sonbull [250]
3 years ago
11

N

Chemistry
2 answers:
bazaltina [42]3 years ago
5 0

Answer:

electron - proton

Explanation:

fsger

blsea [12.9K]3 years ago
3 0

Answer:

A. Electron clouds with a total of seven electrons

Explanation:

A neutral atom of nitrogen must have a total number of 7 electrons in its cloud.

This is the required number of electrons to bring nitrogen into its state of neutrality.

  • To maintain neutrality the number of protons and electrons in an atom must be equal.
  • A charge atom called an ion have either the proton or electron greater.
  • Neutrality is attained when the number of protons and electrons are the same.
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HCl + CaCO3 → CaCl2 + H2O + CO2 balanced equation<br><br><br> Pls Help i will give 100 Points!!!
vagabundo [1.1K]

Answer:

i think its CaCO3 + 2HCl → CaCl2 + H2O + CO2

Explanation:

3 0
2 years ago
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In the following reaction the iodide ion (I-) in HI is oxidized by the permanganate ion (MnO4-). A 5.0 mL solution of 0.20M HI r
katrin [286]

Answer:

The molarity of the KMnO4 solution is 0.05M (option 1)

Explanation:

10 HI + 2 KMnO4 + 3 H2SO4 → 5 I2 + 2 MnSO4 + 8 H2O

This is the base.

10 moles of HI need 2 moles of KMnO4, to make 5 moles of I2 (gas)

Our solution of HI is 0,20 M which means 0,2 moles in 1 L of solution but we used 5 mL, so how many moles did we use?

Molarity . volume = Moles

0.2 moles/L  . 0.005L = 0.001 moles

Notice, we had to convert 5 mL into 0.005 L

So, let's go back to begining: 10 moles of HI need 2 moles of KMnO4.

How many moles of salt, do we need for 0.001 moles of HI.

The rule of three is:

10 moles of HI ___ need ___ 2 moles of KMnO4

0.001 moles of HI __ need ___ (0.001 . 2 ) / 10 = 0.0002 moles

This is our quantity of moles, that we need from KMnO4 but this moles are in 4 mL.

Molarity = moles / L but we can also take account molarity as mM / mL (Molarity/1000)

0.0002 moles . 1000 = 0.2 mM

0.2 mM / 4mL = 0.05 M

6 0
3 years ago
During an effusion experiment, a certain number of moles of an unknown gas passed through a tiny hole in 50 seconds. Under the s
mafiozo [28]
50/35 = √x/ √31.9988 g
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3 0
3 years ago
Use the successive ionization energies for this unknown element to identify the family it belongs to.
NNADVOKAT [17]

Answer: Belongs to the group 2A

Explanation:

As you can see, the first two ionization energies are close and low, meaning that this element ionizates easily.

Not only loses easily the first electron, but the second too

To remove the third electron you requiered a huge amount of energy

Now, elements easily ionizable are the ones from group IA, group 2A and transition metals.

The last ones have mixed characteristics in matter of how many electrons you can remove from them, so they are not a family.

Now the question: group I or group II ?

The elements of group I have low  ionization energies  for the first electron but high energies for the second ones.

Being all that said, the unknown element belongs to the Group 2A

4 0
3 years ago
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