Answer:
Here's what I get
Explanation:
I don't know what the notation on page 91 is (COO⁻?), but your diagram should be something like the one below.
The stearate tails are immersed in and pointing to the centre of the green oil glob.
The negative heads are on the surface surrounded by water molecules.
You may have to draw them as COO⁻.
Answer:
Silver.
Explanation:
Hello!
In this case, we can consider the specific heat as the property that we can analyze in order to answer to this question. In such a way, as the specific heat is known as the energy required to modify the temperature of 1 g of the substance by 1 °C, since the masses of all the substances are the same, we can that their specific heats are respectively 0.240, 0.444 and 4.184 J/(g°C), from the equation:
![Q=mCp\Delta T](https://tex.z-dn.net/?f=Q%3DmCp%5CDelta%20T)
We can see that the higher the specific heat (Cp) the lower the change in temperature considering their inversely proportional relationship. However, as 100 J of energy is applied to all the substances, we can see that silver will exhibit the largest temperature change because a higher change is needed to fit with the provided energy.
Best regards.
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
The process is called evaporation
I think
Answer:
New pressure P2 = 4.95 atm
Explanation:
Given:
Old volume V1 = 1.50 L
New volume V2 = 0.50 L
Old pressure P1 = 1.65 atm
Find:
New pressure P2
Computation:
P1V1 = P2V2
So,
(1.50)(1.65) = (0.50)(P2)
New pressure P2 = 4.95 atm