The reason that some of the elements of period three and beyond are steady in spite of not sticking to the octet rule is due to the fact of possessing the tendency of forming large size, and a tendency of making more than four bonds. For example, sulfur, it belongs to period 3 and is big enough to hold six fluorine atoms as can be seen in the molecule SF₆, while the second period of an element like nitrogen may not be big to comprise 6 fluorine atoms.
The existence of unoccupied d orbitals are accessible for bonding for period 3 elements and beyond, the size plays a prime function than the tendency to produce more bonds. Hence, the suggestion of the second friend is correct.
C. Groups 13-16 contain metalloids.
If you look on a periodic table, these usually have a different color.
Answer:
Explanation:
In this case, we have to start with the <u>chemical reaction</u>:
So, if we start with <u>10 mol of cyclohexanol</u> (
) we will obtain 10 mol of cyclohexanol (
). So, we can calculate the grams of cyclohexanol if we<u> calculate the molar mass:</u>
With this value we can calculate the grams:
Now, we have as a product 500 mL of . If we use the <u>density value</u> (0.811 g/mL). We can calculate the grams of product:
Finally, with these values we can calculate the <u>yield</u>:
%= (405.5/820)*100 = 49.45 %
See figure 1
I hope it helps!
First, in order to calculate the specific heat capacity of the metal in help in identifying it, we must find the heat absorbed by the calorimeter using:
Energy = mass * specific heat capacity * change in temperature
Q = 250 * 1.035 * (11.08 - 10)
Q = 279.45 cal/g
Next, we use the same formula for the metal as the heat absorbed by the calorimeter is equal to the heal released by the metal.
-279.45 = 50 * c * (11.08 - 45) [minus sign added as energy released]
c = 0.165
The specific heat capacity of the metal is 0.165 cal/gC
The best option is a. but it is not complete!
Both liquids have different densities, it's true. But they are immiscible liquids and they don't form a solution.
