Answer: CH₃CN and H₂O.
Explanation:
1) The spieces present in a solution may be either the molecules, in case of covalent compounds, or ions, in case of ionic compounds that dissociate (ionize).
2) Both, CH₃CN and H₂O are covalent (polar covalent) substances, so they do not ionize and the spieces in the solution are the molecules per se.
3) In solution, the molecules of H₂O will solvate the molecules of CH₃CN, meaning that H₂O molecules are able to separate the molecules of CH₃N from each other, and so every molecule of CH₃CN will end surrounded by many molecules of H₂O.
This happens because the interaction between the polar molecules of the two different compounds is strong enough to overcome the intermolecular forces between the molecules of the same compound.
Answer:
A. Strong initial heating caused some of the hydrate sample to spatter out of the crucible.
Explanation:
Hi
The percentage of water in the sample is lower than expected.
A. Strong initial heating caused some of the hydrate sample to spatter out of the crucible:
If part of the sample is splashed from the crucible the mass of water detected will be less.
B. The dehydrated sample absorbed moisture after heating:
If the sample absorbs water after heating the percentage of water would be higher than expected.
C. The amount of the hydrate sample used was too small:
Depending on the sample size, different procedures can be chosen for analysis.
D. The crucible was not heated to constant mass before use:
In many occasions the crucible is heated next to the sample and not in previous form.
E. Excess heating caused the dehydration sample to decompose:
If the sample decomposes during heating, the analysis should be discarded.
success with your homework
Answer:
E. None of these
Explanation:
We know, By GAS laws,
PV = NRT, where p- pressure, v- volume, n- number of moles, R- gas constant ,and T- temperature
Now, In the question, the number of moles remains the same as the gas is the same. so n is constant so we can compare n before and after a temperature change.
= 
where P1= 1 atm, P2 = 10 atm, V1= 20 mL, T1= 10°C and T2= 100°C
We don't have to worry about the standard units as they are present equally on both the sides and get cut, same goes for R( gas constant)
So putting values, we get
Cutting, R on both sides and moving contents to the right so that only V2 is left on the left.
∴ V2 = 
∴ V2 = 20mL
What exactly are you asking
