Answer:
Explanation:
The half-life of K-40 (1.3 billion years) is the time it takes for half of it to decay.
After one half-life, half (50 %) of the original amount will remain.
After a second half-life, half of that amount (25 %) will remain, and so on.
We can construct a table as follows:
No. of Fraction
<u>half-lives</u> <u> t/yr </u> <u>Remaining</u>
0 0 1
1 1.3 billion ½
2 2.6 ¼
3 3.9 ⅛
We see that after 2 half-lives, ¼ of the original mass remains.
Conversely, if two half-lives have passed, the original mass must have been four times the mass we have now.
Original mass = 4 × 2.10 g =
Given mass of KNO₃=346g
Molar mass of KNO₃=(39.098)+(14)+(15.99*3)=101.068gmol⁻¹
Volume of Solution=750ml=0.75dm³
Molarity=(mass of solute/molar mass of solute)*(1/volume of sol. in dm³)
=(346/101.068)*(1/0.75)
=4.56 mol dm⁻³
Answer:
HOMO of 1,3-butadiene and LUMO of ethylene
HOMO of ethylene LUMO of 1,3-butadiene
Explanation:
1,3 - butadiene underogoes cycloaddition reaction with ethylene to give cyclohexene.
According to Frontier molecular orbital theory HOMO of 1,3 butadiene and LUMO of ethylene and HOMO of ethylene and LUMO of ethylene underoges (4 + 2) in thermal or photochemical condition.
Answer:
Amount of Ca(NO3)2 produced = 14.02 g
Explanation:
The given reaction can be depicted as follows:
Ca(OH)2 + 2HNO3 → Ca(NO3)2 + 2H2O
Since it is given that HNO3 is in excess, the limiting reactant is Ca(OH)2
Now, Mass of Ca(OH)2 = 6.33 g
Molar mass of Ca(OH)2 = 74 g/mol
Based on the reaction stoichiometry:
1 mole of Ca(OH)2 forms 1 mole of Ca(NO3)2
Therefore, moles of Ca(NO3)2 produced from the moles of Ca(OH)2 reacted = 0.0855 moles
Molar mass of Ca(NO3)2 = 164 g/mol
The hot water will make sugar to dissolve faster, because in the hot water molecules move faster and diffusion goes faster.
