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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.
Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane
<span>The solid lines between N and Mg are actually ionic bonds. N has 5 valence electrons (2 of which are paired). Of the 3 that are unpaired, 2 are part of covalent bonds with adjacent carbon atoms. N accepts an extra electron to complete its octet, but gets a formal charge of -1. This allows for formation of an ionic bond with Mg, which is +2. Two of these charged N atoms therefore neutralize the charge of the central Mg. As for the coordinate (dative) covalent bonds, Mg has empty orbitals - the ionic bonds with the charged N atoms give it only 4/8 possible valence electrons.
The other two N atoms (dotted lines) have a formal charge of 0 since they form three covalent bonds with adjacent carbon atoms, but they still have a lone pair. Therefore, just to improve stability, each of these N atoms can "donate" its lone pair to Mg in order to complete its octet.
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