What you need to do is find 1/8 of 50
you can just divide 50 by 8 to get 6.25
so now you have to find how many days it will take till there are 6.25 grams of iodine left
every 8.1 days its mass is split in half
so start splitting it in half and every time you do, you add 8.1 days
50/2 =25 8.1
25/2 =12.5 + 8.1
12.5/2= 6.25 +8.1
now you have reached 1/8 of the original amount of Iodine-131
so to find how long it took just add 8.1+8.1+8.1
(this is the same as 8.1x3)
which equals 24.3
it will take 24.3 days for Iodine 131 to decay to 1/8 of its original mass.
(good luck on the regent if thats what your studying for :)
Answer: the correct one would be option B.
Explanation:
Answer : The orbitals that are used to form each indicated bond in citric acid is given below as per the attachment.
Answer 1) σ Bond a: C
O
.
Explanation : The sigma bonds in the 'a' position involves carbon atom which undergoes
hybridisation and has
orbitals of carbon and as well as oxygen atoms involved in the bonding.
Answer 2) π Bond a: C has π orbitals and O also has π orbitals.
Explanation : The pi-bond at the 'a'position has carbon and oxygen atoms which undergoes pi-bond formation. And has pi orbitals of oxygen and carbon involved in the bonding process.
Answer 3) Bond b: O
H has only S orbital involved in bonding.
Explanation : The bonding at 'b' position involves oxygen and hydrogen atoms in it. It has
hybridized orbitals and S orbital of hydrogen involved in the bonding.
Answer 4) Bond c: C
O
Explanation : The bonding involved at 'c' position has carbon and oxygen atoms involved in it. Both the atoms involves the orbitals of
hybridized bonds.
Answer 5) Bond d: C
C 
Explanation : The bonding involved at the position of 'd' involves two carbon atoms. Therefore, they undergo
hybridization. The orbitals involved in this hybridization are also
.
Answer 6) Bond e: C1 containing O
C2 
Explanation : The bonding at the 'e' position involves two carbon atoms on is containing oxygen with double bonds and the C2 carbon atom. The carbon containing oxygen has
hybridized orbitals involved in the bonding; whereas C2 carbon ha
orbitals.
Answer:
0.0222 mole of NaOH is needed to react with NH4F
Explanation:
NH4F + NaOH --> NaF + NH3 + H2O
Data given
Mass of NH4F =0.821g, Concentration of NaOH= 1M, volume of NaoH =25ml
But mole = (CV)/1000
given mole of NaoH = (1 * 25)/1000 = 0.025moles of NaOH used
Molar mass of NH4F = 37g/mol
mole of NH4F used 0.821 / 37 = 0.0222 mole NH4F
Determine the excess and limiting reactant,
NaOH is in excess
0.025 - 0.0222 = 0.0028 mole NaOH excess
0.0222 mole of NaOH is required to react with NH4F
The difference of the sample mass and the heated sample mass is the amount of water which evaporated by heating process, then we can calculate:
mass of H₂O = 30.7 - 22.9 = 7.8 g
then calculate the number of moles:
number of moles of H₂O = mass / molecular mass
number of moles of H₂O = 7.8 / 18 = 0.43 moles
So the hydration number is 0.43.
Sr(NO₃)₂ • 0.43 H₂O