Answer:
40g
Explanation:
Solubility of Copper sulfate at 90°=60g
Solubility of potassium bromide at 90°=100g
100g-60g=40g
Answer:
(a) 91 kg (2 s.f.) (b) 22 m
Explanation:
Since it is stated that a constant horizontal force is applied to the block of ice, we know that the block of ice travels with a constant acceleration and but not with a constant velocity.
(a)

Subsequently,

*Note that the equations used above assume constant acceleration is being applied to the system. However, in the case of non-uniform motion, these equations will no longer be valid and in turn, calculus will be used to analyze such motions.
(b) To find the final velocity of the ice block at the end of the first 5 seconds,

According to Newton's First Law which states objects will remain at rest
or in uniform motion (moving at constant velocity) unless acted upon by
an external force. Hence, the block of ice by the end of the first 5
seconds, experiences no acceleration (a = 0) but travels with a constant
velocity of 4.4
.

Therefore, the ice block traveled 22 m in the next 5 seconds after the
worker stops pushing it.
The average rate of change of distance over the time interval
3 ≤ t ≤ 6 represents the coin's average velocity over that interval.
Answer:
(a) 6.567 * 10^15 rev/s or hertz
(b) 8.21 * 10^14 rev/s or hertz
Explanation:
Fn= 4π^2k^2e^4m * z^2/(h^3*n^3)
Where Fn is frequency at all levels of n.
Z = 1 (nucleus)
e = 1.6 * 10^-19c
m = 9.1 * 10^-31 kg
h = 6.62 * 10-34
K = 9 * 10^9 Nm2/c2
(a) for groundstate n = 1
Fn = 4 * π^2 * (9*10^9)^2*(1.6*10^-19)^4* (9.1 * 10^-31) * 1 / (6.62 * 10^-31)^3 = 6.567 * 10^15 rev/s
(b) first excited state
n = 1
We multiple the groundstate answer by 1/n^3
6.567 * 10^15 rev/s/ 2^3
F2 = 8.2 * 10^ 14 rev/s