The volume of a gas that is required yo react with 4.03 g mg at STP is 1856 ml
calculation/
- calculate the moles of Mg used
moles=mass/molar mass
moles of Mg is therefore=4.03 g/ 24.3 g/mol=0.1658 moles
- by use of mole ratio of Mg:O2 from the equation which is 2:1
the moles 02=0.1679 x1/20.0829 moles
- at STP 1 mole of a gas= 22.4 l
0.0895 moles=? L
- =0.0895 moles x22.4 l/ 1 mole=1.8570 L
into Ml = 1.8570 x1000=1856 ml approximately to 1860
卂几丂山乇尺
As the name indicate that in 1-octene there is a double bond on the first position in the eight carbon atoms chain THUS to detect its presence, we use bromine water test;
Add bromine water to the test tube containing octane/octene if the brown colour of bromine water disappears then octene is present.
Because more than one substance was released (following a color change signifying a chemical reaction), the sample was indeed, a compound.
If this is a single atom of Boron, there should be 5 electrons as well. Boron as an electron configuration of 2-3 or 1s2 2s2 2p
Answer: (2)3.0 mol
Explanation: 222.3g= 222.3/74= 3.0 mol
1mol Ca(OH)2 is 74g
