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3 years ago

How many valence electrons should you use to draw a Lewis structure of ammonia (NH3)?

Chemistry

2 answers:

stealth61 [152]3 years ago

7 0

You should use 8 valence electrons to draw a Lewis structure of ammonia. b) 8

Mekhanik [1.2K]3 years ago

4 0

Hello there,

Your correct answer would be 8. There are 8 valence electrons available for the Lewis structure for NH3.

Hope this helps

~Hottwizzlers

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In this double replacement reaction, which are the products? hcl + naoh —> ______?

Anarel [89]

The answer is: " NaCl + H₂O " ; (or; write as: " H₂O + NaCl " ) .
________________________________________________________
Specifically:
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    HCl + NaOH —> NaCl + H₂O ;   or; write as:

   NaOH + HCl —> H₂O + NaCl .
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            This type of "double-replacement" reaction is called "neutralization".

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3 years ago

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MrRissso [65]

Answer:

A is a physical change, no atoms are being lost its still in it original form

B is a chemical change, it is losing atoms and changing into a new substance

Explanation:

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3 years ago

A neutral atom has two, eight, eight electrons in its first, second, and third energy levels. this information ________. does no

olga55 [171]

This info tells us if it stable or not and it tells us how many electrons it need to be stable.

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3 years ago

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Halp!

Nadya [2.5K]

Answer:

One with mass 0.52kg will have higher acceleration.

Explanation:

We know that

F = m x a

As we can see, for a constant value of F , for a body having more mass will have less acceleration and a body having less mass will have more acceleration ( Inversely proportional when force applied is constant).

Therefore lesser mass body will have more acceleration.

Hence, 0.52 kg body will have more acceleration.

Verification :

F = ma

For 0.52kg:

a= F / m = 26 / 0.52 = 50 m/s²

For 0.78kg:

a= F/m = 26 / 0.78 = 33.3 m/s²

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3 years ago

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ahrayia [7]

Radioactive elements undergoes first order decay.

Now, for 1st order decay kinetics, rate constant (k) can be estimated as follows

k = 0.639/ t1/2,
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Also, we know that
$t = \frac{2.303}{k}log \frac{Co}{Ct}$ .................(1)

Here, Co = initial conc. of radioactive element
Ct = conc. of radioactive element at time t
t = time required to reach the concentration Ct

Thus, equation 1 can be used to estimate the amount of lead-209 present after 't' days.

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