It is covalent bonding. The electrons are shared between the phosphorus and the chlorines.
covalent bonding is when electrons are shared between two elements.
molecular polarity is a little bit complicated, but I will try to explain ;)
PCl3 is an alternation on tetrahedral molecules.
It means that P has one lone pair of electrons. This pair of electrons are only attracted to the P nuclei and thus a greater freedom of motion.
This means that their orbital is bigger and this pushes the 3 Cl atoms closer together.
The angle between each Cl now is 107 and the angle between Cls and P is greater than 107.
Now, due to this shape, and also electronegativity (Cl is more electronegative than P meaning that it tends to hog the electrons they share closer to itself), PCl3 is polar. Electrons that are shared tend to flow closer towards the Cl than the P side.
Therefore, on the Cl side of the molecule it's, more negative. On the P side, it's more positive.
<span>2.40 - 1.68 =0.72 g of oxigen
moles = 0.72/16 g/mol=0.045
moles x = 1.68/ 55.9=0.03
0.03/0.03 = 1 = x
0.045 / 0.03 = 1.5 = O
to get whole numbers multiply by 2
x2O3
X2O3 +3 CO = 2 X + 3 CO2</span>
Hello.
The answer is C.Amine
When an amine is combined (reacted) with a carboxyl group, an AMIDE + water is formed, and if you carry on heating under a vacuum, an imidazoline is formed.
Have a nice day
Answer:
4.00 is the pH of the mixture
Explanation:
The ethyl amine reacts with HNO3 as follows:
C2H5NH2 + HNO3 → C2H5NH3⁺ + NO3⁻
To solve this question we need to find the moles of ethyl amine and the moles of HNO3:
<em>Moles C2H5NH2:</em>
0.0500L * (0.100mol/L) = 0.00500 moles ethyl amine
<em>Moles HNO3:</em>
0.201L * (0.025mol/L) = 0.005025 moles HNO3
That means HNO3 is in excess. The moles in excess are:
0.005025 moles HNO3 - 0.00500 moles ethyl amine =
2.5x10⁻⁵ moles HNO₃
In 50 + 201mL = 251mL = 0.251L:
2.5x10⁻⁵ moles HNO₃ / 0.251L = 9.96x10⁻⁵M = [H+]
As pH = -log [H+]
pH = -log 9.96x10⁻⁵M
pH = 4.00 is the pH of the mixture
Answer:
1
Explanation:
1 hydrogen is displaced from H2so4 in the reaction
