Percent yield or yield is mathematically defined as:
Yield = Actual amount / Theoretical amount
So to solve the yield, let us first calculate the
theoretical amount of POCl3 produced. The balanced chemical reaction for this
is:
6 PCl5 + P4O10 ---> 10 POCl3
Since P4O10 is stated to be supplied in large amount, then
PCl5 becomes the limiting reactant.
So we calculate for POCl3 based on PCl5. To do this let us
convert the amount into moles: (molar mass PCl5 = 208.24 g/mol)
n PCl5 = 42.66
grams / (208.24 g/mol)
n PCl5 = 0.205 mol
Now based on the
stoichiometric ratio of the reaction:
<span>n POCl3 = 0.205 mol (10
POCl3 / 6 PCl5)
n POCl3 = 0.3414 mol POCl3</span>
Converting to mass (molar
mass POCl3 = 153.33 g/mol)
m POCl3 = 0.3414 mol (153.33 g/mol)
m POCl3 = 52.35 g
Calculating for yield:
Yield = 47.22 g/ 52.35 g
Yield = 0.902
<span>%Yield = 90.2 % (ANSWER)</span>
Generally when we move down the group on a periodic table the atomic radii increases as the valency electrons occupy higher levels due to the increasing quantum number. Hence the atomic radii increases down the group.
The ionic radii increases down the group because while we move down the group the elements gain electrons and form ions called anions as an additional electron occupies the orbital the ions get bigger in size. Hence the ionic radii increase.
Electronegativity is described as the ability to attract and bind with electrons and it is a qualitative property. It decreases as we move down the group because the distance between the valency electrons and the nucleus increases. Hence electronegativity decreases down the group.
Reactivity increases as we move down the group as the metals have the tendency to lose electron form its outer shell.
Therefore the answer is ionic radii increases.
Answer:
b. ICl experiences dipole-dipole interactions
Explanation:
The molecule with the <em>stronger intermolecular forces</em> will have the higher boiling point.
In Br₂, the Br-Br bond has a <em>dipole moment of zero</em>, because the two atoms are identical.
In ICl, the I-Cl bond, has two different atoms. One must be more electronegative than the other, so there will be a <em>non-zero bond dipole</em>.
ICl will have the higher boiling point.
a is <em>wrong</em>. Br₂ is nonpolar, so it has no dipole-dipole interactions.
c is <em>wrong</em>. Br₂ cannot form hydrogen bonds, because there is no hydrogen.
d is <em>wrong</em>. ICl has dipole-dipole interactions.
Hello!
The reaction between HBr and KOH is the following:
HBr+KOH
→H₂O + KBr
To calculate the amount of HBr left after addition of KOH, you'll use the following equations:
![HBr_f=HBr_i-KOH=([HBr]*vHBr)-([KOH]*vKOH) \\ \\ HBr_f=(0,25M*0,64L)-(0,5M*0,32L)=0 mol HBr](https://tex.z-dn.net/?f=HBr_f%3DHBr_i-KOH%3D%28%5BHBr%5D%2AvHBr%29-%28%5BKOH%5D%2AvKOH%29%20%5C%5C%20%20%5C%5C%20HBr_f%3D%280%2C25M%2A0%2C64L%29-%280%2C5M%2A0%2C32L%29%3D0%20mol%20HBr)
That means that after the addition of 32 mL of KOH, there is no HBr left in the solution and the pH should be
neutral, close to 7.
Have a nice day!
Answer:
A lithium atom has 3 protons and 3 electrons. It can lose one of its electrons, making it an ion. It now has more positive protons than electrons so it has an overall positive charge. Therefore it is a positive ion.
Explanation:
Lithium, an alkali metal with three electrons, is also an exception to the octet rule. Lithium tends to lose one electron to take on the electron configuration of the nearest noble gas, helium, leaving it with two valence electrons.
