Water has h bonding
H-H
Sodium fluoride
I think
Answer:
empirical formula = C3H7
molecular formula = C6H14
Answer:
a. -0.63 V
b. No
Explanation:
Step 1: Given data
- Standard reduction potential of the anode (E°red): -1.33 V
- Minimum standard cell potential (E°cell): 0.70 V
Step 2: Calculate the required standard reduction potential of the cathode
The galvanic cell must provide at least 0.70V of electrical power, that is:
E°cell > 0.70 V [1]
We can calculate the standard reduction potential of the cathode (E°cat) using the following expression.
E°cell = E°cat - E°an [2]
If we combine [1] and [2], we get,
E°cat - E°an > 0.70 V
E°cat > 0.70 V + E°an
E°cat > 0.70 V + (-1.33 V)
E°cat > -0.63 V
The minimum E°cat is -0.63 V and there is no maximum E°cat.
Okay, so even if I just gave you the answers, your teacher needs work on it too so it'll be easier/better if I just explain how to do it.
Basically, both sides need to have the same number of molecules. To do this, we make charts. This is the first side of number one:
Na - 1
Mg- 1
F - 2
The subscript gives F two molecules, and the other ones only each have one. This is the second side:
Na- 1
Mg- 1
F- 1
So they're not equal. To fix this, we add coefficients. These are numbers that are going to appear in the front of each compound/element and changes the number of molecules of the WHOLE compound/element. We need two F on the second side, so we'll put a coefficient of 2 in front of NaF. The new chart for the second side is this:
Na- 2
Mg- 1
F- 2
Now we've fixed the F, but now Na is off! So let's go to the first side again and see what we can do. We can put a 2 in front of the Na. The new chart is this:
Na- 2
Mg -1
F- 2
Now both sides are the same. The full new equation is:
2Na + MgF(sub2) = 2NaF + Mg
Basically, do this for all of them. Feel free to ask more questions.
Answer: The final temperature of both the weight and the water at thermal equilibrium is 
.
Explanation:
The given data is as follows.
mass = 7.62 g, 
Let us assume that T be the final temperature. Therefore, heat lost by water is calculated as follows.
q = 
= 
Now, heat gained by lead will be calculated as follows.
q = 
= 
According to the given situation,
Heat lost = Heat gained
=
T =
Thus, we can conclude that the final temperature of both the weight and the water at thermal equilibrium is .
