Is A have a nice day good luck b
These oppositely charged compounds are strongly held by electrostatic forces of attraction as these be together for a long time the rise in temparature occurs so that the the melting points rises in them.
Answer:
C = (5/9) F - (160/9)
They both read equal at Z = - 40
Explanation:
We are looking for a linear function so we can write the following condition
Y = aX + b
Applying it to the exercise we got C = a F + b
Let's use the facts that C = 0 when F = 32 and C = 100 when F = 212
0 = 32 a + b (1)
100 = 212 a + b (2)
From (1) b = - 32 a , when we replace this in (2) we obtain a = (5/9)
and b = - (5/9)32 = - 160/9
Finally the linear function is C = (5/9) F - (160/9)
Both readings are equal at a Z number so
Z = (5/9) Z - 160/9
(4/9) Z = -160/9 and Z = - 40
A stable isotope has just<em> the right number of neutrons for the number of protons </em>(the <em>n:p ratio</em>) to hold the nucleus together against the repulsions of the protons.
A radioactive isotope has either too few or too many neutrons for the nucleus to be stable,
The nucleus will then emit <em>alpha, beta, or gamma radiation</em> in an attempt to become more stable.
Answer:
mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g
Explanation:
The partition coefficient of X between ethoxy ethane (ether) and water, K is given by the formula
K = concentration of X in ether/concentration of X in water
Partition coefficient, K(X) between ethoxy ethane and water = 40
Concentration of X in ether = mass(g)/volume(dm³)
Mass of X in ether = m g
Volume of ether = 50/1000 dm³ = 0.05 dm³
Concentration of X in ether = (m/0.05) g/dm³
Concentration of X in water = mass(g)/volume(dm³)
Mass of X in water left after extraction with ether = (5 - m) g
Volume of water = 1 dm³
Concentration of X in water = (5 - m/1) g/dm³
Using K = concentration of X in ether/concentration of X in water;
40 = (m/0.05)/(5 - m)
(m/0.05) = 40 × (5 - m)
(m/0.05) = 200 - 40m
m = 0.05 × (200 - 40m)
m = 10 - 2m
3m = 10
m = 10/3
m = 3.33 g of X
Therefore, mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g
