<span>Carboxylic acid.
This is because there is no reducing termini. and if one end is a methyl, that is just hydrocarbons, and the rest must also be hydrocarbons. We can assume that there are no other constituents that are making functional groups and given with what we have this is a carboxylic acid.</span>
Answer:
A
Explanation:
PV=nRT
PV/nT
V/T -> (1)/(300)=(x)/(150)
x=.420
Answer:
ΔH = -976.5 kJ
Explanation:
For the reaction given, there are 2 moles of benzene (C6H6). The heat of this reaction is -6278 kJ, which means that the combustion of 2 moles of benzene will lose 6278 kJ of heat. It is an exothermic reaction.
The value of ΔH, the enthalpy, is a way of measurement of the heat, and it depends on the quantity of the matter (number of moles).
So, 24.3 g of benzene has :
n = mass/ molar mass
n = 24.3/78.11
n = 0.311 moles
2 moles ------------ -6278 kJ
0.311 moles ----------- x
By a simple direct three rule:
2x = -1953.08
x = -976.5 kJ
Answer:
The number of periods shows you the number of _______
<u>Electron Shell</u>
<u></u>
Explanation:
<u><em>Modern periodic table is divided into Groups and Periods.</em></u>
A period is the horizontal row of the periodic table. There are <u>seven periods </u>in the periodic table.
It describes the following :
- The period in which a particular element found will show its <u>electron shell</u>
- The first period has Hydrogen and Helium = 1 - shell
- The second Period has Lithium to neon = 2- shells
A group is the vertical column of the periodic table . There are <u>18 groups</u> in total.
It describes the following :
The group number of the element is same as the<u> number of electrons in the outermost shell.</u>
The element having group 1 has = 1 electron in outer shell (example Na, K)
The element having group 2 has = 2 electron in outer shell (example Ca, Mg)
The element having group 0 has = full outer shell (example Ne Xe). these are also known as noble gases
The amount of HCl required for one experiment - 13.5 µl
the volume in terms of L - 13.5 x 10⁻⁶ L
the volume of HCl available - 0.250 L
since one experiment uses up - 13.5 x 10⁻⁶ L
then number of experiments - 0.250 L / 13.5 x 10⁻⁶ L = 1.8 x 10⁴ times
the experiment can be carried out 18000 times
