Answer:- The hydroxide ion concentration of the solution is 
.
Solution:- The formula used to calculate pOH from hydroxide ion is:
![pOH=-log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-log%5BOH%5E-%5D)
When pOH is given and we are asked to calculate hydroxide ion concentration then we multiply both sides by negative sign and take antilog and what we get on doing this is:
![[OH^-]=10^-^p^O^H](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E-%5Ep%5EO%5EH)
pOH is given as 5.71 and we are asked to calculate hydrogen ion concentration. Let's plug in the given value in the formula:
![[OH^-]=10^-^5^.^7^1](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E-%5E5%5E.%5E7%5E1)
![[OH^-]](https://tex.z-dn.net/?f=%5BOH%5E-%5D)
= 0.00000195 or
So, the hydroxide ion concentration of the solution is .
Velocity and mass are directly proportional to the quantity of momentum by:
p = mv. Therefore, and increase in either velocity or mass will lead to an increase in momentum and vice versa. Momentum during a reaction is always conserved, meaning that the mass and initial velocity before a reaction will always be equal to the change in mass and velocity produced after the reaction. Kinetic energy after a reaction, however, is not always conserved. For example if a fast moving vehicle collided with a stationary vehicle, and moved together, the overall kinetic energy would be after the reaction, as a heaver mass would be moved by the same velocity causing a decrease in kinetic energy.
I don't know if this is exactly what you are looking for, but in physics this is how it is understood.
Answer:
D is correct
Explanation:
because
we know that
density of lead is 11.36 g/cm3
and
density of tin is 7.31 g/cm3
so..
density of alloy by mixing 50/50
=(11.36+7.31)/2 g/cm3
=18.67/2 g/cm3
=9.33 g/cm3
Given:
Be - Beryllium - 9,3227
C - Carbon - 11,2603
O - Oxygen - 13,6181
Ne - Neon - 21,5645
B - Boron - 8,298
Li - Lithium - 5,3917
F - Fluorine - 17,4228
N - Nitrogen - 14,5341
Arranged from highest ionization energy to lowest ionization energy.
Ne ; F ; N ; O ; C ; Be ; B ; Li
