To bond with another atom
Answer:
BRO WHAT THEY ANSWER TO UR BUT NOT MINE OK
Explanation: I THINK THE ASNSWER IS C BUT NOT SURE.... HOPE IT HELPS
Answer:
Explanation:
C + O2 → CO2
Mole of C = 24 g/(12 g/mole)
Mole of C = 2 mole
Mole of molecular O2 = 74 g/(32 g/mole)
Mole of molecular O2 = 2.3125 mole
Since mole of C < mole of O2, then C being the limiting reagent.
From the reaction, it shows that mole ratio between C and O2 = 1 : 1.
So, 2 moles of C will stoichiometrically react with 2 moles of O2 to generate 2 moles of CO2.
Avogadro's law states that :"equal volumes of all gases, at the same temperature and pressure, have the same number of molecules i.e. 6.02 x 10^23 molecules/mole.
Therefore, 2 moles of CO2 contain 2 moles x 6.02 x 10^23 molecules/mole = 1.204 x 10^24 molecules of CO2 is formed.
Answer:
<u>One lone-Pair is present in Ammonia</u>
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Explanation:
The number of valence electron in N = 5
The number of Valence electron in H = 1
The formula of ammonia = NH3
Total valence electron in ammonia molecule = 5 +3(1) = 5+3 = 8
The lewis structure suggest that :
Nitrogen completes its octet by sharing the electron pair with 3 hydrogen atoms.
3 electron of Nitrogen are involved in sharing with Hydrogen
So,<u><em> remaining two electron are left non-bonded</em></u> . Hence they exist as lone- pair
So, there is only 1 lone pair in the ammonia molecule .
The shape of NH3 is bent according to VSEPR theory . This is so because the presence of 1 lone pair causes more repulsion and occupy more space.
Thus the lone pair is changing the shape of the ammonia molecule . It also increase the dipole moment of the molecule , which gives polarity to it.
<span>A lahar is a type of mudflow that occurs after a volcanic eruption. It is a sort of mudflow or trash stream made out of a slurry of pyroclastic material, rough flotsam and jetsam, and water. The material streams down from a fountain of liquid magma, normally along a waterway valley. Lahar is the most deadly by-products of the eruption because of the speed they can travel. </span>
