She would observe a yellowish solid precipitate which is the lead iodide and a white solid precipitate which is the potassium nitrate.
This is because the lead nitrate solution which contains particles of lead will mix with the potassium iodide solution containing particles of iodide. Upon mixing,the lead particles from the Lead nitrate solution combines with the iodide particles from Potassium iodide and form two compounds, a yellowish solid precipitate called lead iodide and a white solid precipitate called Potassium nitrate.
The formation of entirely two new compounds is known as the double displacement reaction and can be written in a chemical equation as
2KI(aq.)+Pb(NO₃)₂(aq.)------>2KNO₃(aq.)+PbI ₂(s)
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Answer:
25.30 gram
Explanation:
No of moles = given mass / molar mass
No of moles = 63.17/80.06
0.7890 moles
Mass of sulphar = no of moles× molar mass of sulphar
Mass of sulphur = 0.7890×32.065
25.30 gram
Answer:
Explanation:
A tertiary alcohol is a compound (an alcohol) in which the carbon atom that has the hydroxyl group (-OH) is also bonded (saturated) to three different carbon atoms.
Based on the question, the only <u>tertiary alcohol that can result from C₆H₁₄O that have a 4-carbon chain</u> is
2-hydroxy-2,3-dimethylbutane
H OH H H
| | | |
H - C - C - C - C - H
| | | |
H CH₃ CH₃ H
From the above, we can see that the carbon atom having the hydroxyl group is also bonded to three other carbon atoms. And since we aren't considering stereochemistry, this is the only tertiary alcohol we can have with a 4-carbon chain
(a) In this section, give your answers to three decimal places.
(i)
Calculate the mass of carbon present in 0.352 g of CO
2
.
Use this value to calculate the amount, in moles, of carbon atoms present in 0.240 g
of
A
.
(ii)
Calculate the mass of hydrogen present in 0.144 g of H
2
O.
Use this value to calculate the amount, in moles, of hydrogen atoms present in 0.240 g
of
A
.
(iii)
Use your answers to calculate the mass of oxygen present in 0.240 g of
A
Use this value to calculate the amount, in moles, of oxygen atoms present in 0.240 g
of
A
(b)
Use your answers to
(a)
to calculate the empirical formula of
A
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