A homogenous mixture is uniform and thus hard to recognize as a mixture. An example is water.
The volume becomes two. You have to use the equation P1 x V1 = P2 x V2
P is pressure and V is volume.
P1 = 50 P2 = 125
V1 = 5 V2 = v (we don't know what it is)
Then set up the equation:
50 times 5 = 125 times v
250 = 125v
the divide both sides by 125 and isolate v
2 = v
Therefore the volume is decreased to 2.
Also, Boyle's Law explains this too: Volume and pressure are inversely related, This means that when one goes up the other goes down (ie when pressure increases volume decreases and vice versa). Becuase the pressure went up from 50 KPa tp 125 KPa the volume had to decrease.
Answer:
9.36
Explanation:
Sodium formate is the conjugate base of formic acid.
Also,
for sodium formate is 
Given that:
of formic acid = 
And, 
So,
Concentration = 0.35 M
HCOONa ⇒ Na⁺ + HCOO⁻
Consider the ICE take for the formate ion as:
HCOO⁻ + H₂O ⇄ HCOOH + OH⁻
At t=0 0.35 - -
At t =equilibrium (0.35-x) x x
The expression for dissociation constant of sodium formate is:
![K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}](https://tex.z-dn.net/?f=K_%7Bb%7D%3D%5Cfrac%20%7B%5BOH%5E-%5D%5BHCOOH%5D%7D%7B%5BHCOO%5E-%5D%7D)
Solving for x, we get:
x = 0.44×10⁻⁵ M
pOH = -log[OH⁻] = -log(0.44×10⁻⁵) = 4.64
pH + pOH = 14
So,
<u>pH = 14 - 4.64 = 9.36</u>
Explanation:
(A)role of nittogen fixing bacteria
=Nitrogen-fixing bacteria, microorganisms capable of transforming atmospheric nitrogen into fixed nitrogen (inorganic compounds usable by plants). More than 90 percent of all nitrogen fixation is effected by these organisms, which thus play an important role in the nitrogen cycle.
B)role of nitrifying bacteria
=Nitrifying bacteria convert the most reduced form of soil nitrogen, ammonia, into its most oxidized form, nitrate. In itself, this is important for soil ecosystem function, in controlling losses of soil nitrogen through leaching and denitrification of nitrate.
C)role of denitrifying bacteria
=Denitrifying bacteria converts nitrates back to nitrogen gas.
Answer:
Option D. KBr < KCl < NaCl
Explanation:
We'll begin by calculating the number of mole of each sample.
This can be obtained as follow:
For NaCl:
Mass = 1 g
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mole of NaCl =?
Mole = mass /Molar mass
Mole of NaCl = 1/58.5
Mole of NaCl = 0.0171 mole
For Kbr:
Mass = 1 g
Molar mass of KBr = 39 + 80 = 119 g/mol
Mole of KBr =?
Mole = mass /Molar mass
Mole of KBr = 1/119
Mole of KBr = 0.0084 mole
For KCl:
Mass = 1 g
Molar mass of KCl = 39 + 35.5 = 74.5 g/mol
Mole of KCl =?
Mole = mass /Molar mass
Mole of KCl = 1/74.5
Mole of KCl = 0.0134 mole
Summary
Sample >>>>>>>> Number of mole
NaCl >>>>>>>>>> 0.0171
KBr >>>>>>>>>>> 0.0084
KCl >>>>>>>>>>> 0.0134
Arranging the number of mole of the sampl in increasing order, we have:
KBr < KCl < NaCl
