I believe a solution of Sn(NO3)2 can not be stored in an aluminium container because Aluminium is higher in the reactivity series compared to Tin (Sn). Therefore, Aluminium is more reactive than Tin and hence aluminium will displace Tin from its salt forming Aluminium nitrate and Tin metal. Thus storing Tin nitrate in an aluminium container will cause the "eating away' of the container.
Answer:
Aldehyde and ketone both react with alcohols (2 equivalent) in the presence of acid catalyst to give acetals.
Explanation:
Alcohols are poor nucleophiles and so protonation of the carbonyl oxygen used to make the carbonyl carbon a stronger electrophile. Addition of the first equivalent of alcohols gives a hemiacetal,
Addition of the second equivalent of alcohol is accompanied by loss of one molecule water to yield the produce acetal.
Answer:
Square planar.
Explanation:
Hello,
In this case, it can be demonstrated that the central atom Xenon in XeF₄ has a sp³d² hybridization which means that its geometry is likely to be octahedral. Nevertheless Applying by applying the valence shell electron pair repulsion (VSEPR) model to in order to minimize the repulsion among lone pairs, bond pairs and lone pair-bond pairs, we we realize repulsion is actually minimized when lone pairs are anti to one another and the fluorine atoms are in equatorial position, for that reason, the corrected and properly exhibited geometry or molecular arrangement of the compound turns out square planar. You can verify it on the attached picture.
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Answer:
36°C
Explanation:
Given parameters:
Mass of aluminum = 725g
Quantity of heat = 2.35 x 10⁴J
Unknown:
Temperature change = ?
Solution:
To solve this problem, we simply use the expression below:
The quantity of energy is given as:
Q = m C Δt
Q is the quantity of energy
m is the mass
C is the specific heat capacity of aluminum = 0.9J/g°C
Δt is the change in temperature
The unknown is Δt;
Δt =
=
= 36°C
Answer:
Samples A and B are mixtures
Explanation:
For sample A, we were told that the initial density of the solution is 0.77/gmL. After distillation, the liquid left has a density of 1.04/gmL. Apparently, A is a mixture of liquids.
As for sample B, the spread of the chromatogram resulting in different colours show that the sample B is not pure. It is a mixture of substances. A pure sample will yield only a single spot.