alpha decay is an emission of ₂⁴He
nucleus. If an element undergoes an alpha decay, the mass of the daughter
nucleus formed is reduced by 4 compared to mass of parent atom and atomic
number is reduced by 2 compared to atomic number of parent atom.
Let's assume that the parent atom is X, atomic number is z and mass number is A. Then the formed daughter nucleus, Y should have atomic number as z-2 and mass number as A-4.
The equation is
(z) (A)X → ₂⁴α + (z-2)(A-4)Y
Example:
₉₄²³⁹Pu → ₂⁴α + ₉₂<span>²³⁵U</span>
Because in difrent materials atoms are more compact or less compact.if they are less compact then it will be easear for them to move
The reaction is:
C₂H₄ + 3O₂ → 2CO₂ + 2H₂O; ΔH = 1410 kJ
When we reverse this reaction, the sign of the enthalpy change, ΔH, will be changed. The enthalpy change for the reversed reaction would be 1,410 kJ.
Next, we must also multiply the reaction by 2, so the final enthalpy change for the reverse reaction will be:
ΔH = 2,820 kJ
Answer:
Saturated fat occurs naturally in red meat and dairy products. It's also found in baked goods and fried foods. Trans fat occurs naturally in small amounts in red meat and dairy products. Trans fat can also be manufactured by adding hydrogen to vegetable oil.
Answer:
N-ethyl-2-methylpropan-2-amine
Explanation:
In this case, we have to start with the <u>IR info</u>. The signal on 3400 cm^-1 indicates the presence of a <u>hydrogen bonded to the heteroatom</u>. In this case, we have nitrogen in the formula, so we will have the <u>amine group</u>.
On the other hand, we have to analyze the NMR info:
a) We have 2 singlets => This indicates the presence of 2 different hydrogens without neighbors.
b) We have a triplet => This indicates the presence of <u>CH3 bonded to a CH2</u>.
c) We have a quartet => This indicates the presence of <u>CH2 bonded to a CH3</u>.
From b) and c) we can conclude that we have the <u>ethyl group</u> bonded to a nitrogen.
Finally, we have to add 4 more carbons in such a way that we only have a single signal. In this case the <u>ter-butyl group</u>.
In that way, we will have <u>2 singlets</u> (from the CH3 groups in the ter-butyl and the H on the N). Also, we will have the <u>quartet </u>on the CH2 in the ethyl group and the <u>triplet</u> on the CH3 in the ethyl group
