1.For each of the following, give the name of an element from Period 4 (potassium to krypton), which matches the description.
Elements may be used once, more than once or not all.. Single line text.
(1 Point)
an element that reacts with water to produce a lilac flame
2.For each of the following, give the name of an element from Period 4 (potassium to krypton), which matches the description.
Elements may be used once, more than once or not all.. Single line text.
(1 Point)
an element used as an inert atmosphere
3.For each of the following, give the name of an element from Period 4 (potassium to krypton), which matches the description.
Elements may be used once, more than once or not all.. Single line text.
(1 Point)
an element that has a valency of 3
4.Write a balanced chemical equation for the reaction between potassium and water. (Non-anonymous question). .
(1 Point)
Upload file
File number limit: 1Single file size limit: 10MBAllowed file types: Word, Excel, PPT, PDF, Image, Video, Audio
5.For each of the following, give the name of an element from Period 4 (potassium to krypton), which matches the description.
Elements may be used once, more than once or not all.. Single line text.
(1 Point)
an element with a fixed valency of 2 that not is not in group 2
Help
<u>Answer:</u> The value of <em>i</em> is 1.4 and 40% dissociation of 100 particles of zinc sulfate will yield 60 undissociated particles.
<u>Explanation:</u>
The equation used to calculate the Vant' Hoff factor in dissociation follows:
where,
= degree of dissociation = 40% = 0.40
i = Vant' Hoff factor
n = number of ions dissociated = 2
Putting values in above equation, we get:
The equation used to calculate the degee of dissociation follows:
Total number of particles taken = 100
Degree of dissociation = 40% = 0.40
Putting values in above equation, we get:
This means that 40 particles are dissociated and 60 particles remain undissociated in the solution.
Hence, 40% dissociation of 100 particles of zinc sulfate will yield 60 undissociated particles.
Answer:
51 J
Explanation:
The air inside a bicycle tire pump has 27 joules of heat conducted away. By convention, when heat is released, it takes the negative sign, so Q = -27 J.
77.9 joules of work done are being done on the air inside a bicycle tire pump. By convention, when work is being done on the system, it takes the positive sign, so W = 77.9 J
We can calculate the change in the internal energy (ΔU) using the following expression.
ΔU = Q + W
ΔU = (-27 J) + 77.9 J
ΔU = 51 J
