The number of potassium atom that are in 0.25 moles potassium carbonate is calculated as follows
by use of Avogadro contant
1 mole= 6.02 x10^23 atoms
what about 0.25 moles,
by close multiplication
{0.250 moles x 6.02 x10^23} / 1 mole = 1.505 x10^23 atoms
<span>the pH of a 0.050 M triethylamine, is 11.70
</span>
For triehtylamine,
, the reaction will be
and we know, pH = -log[H+] and pOH = -log[OH-]
Also, pOH + pH = 14
Now, the Kb value = 5.3 x 10^-4
And
![kb = \frac{( [( C_{2}H_{5})_{3}NH^{+} ]* OH^{-} )}{[( C_{2}H_{5})_{3}N]}](https://tex.z-dn.net/?f=kb%20%3D%20%20%5Cfrac%7B%28%20%5B%28%20C_%7B2%7DH_%7B5%7D%29_%7B3%7DNH%5E%7B%2B%7D%20%5D%2A%20%20OH%5E%7B-%7D%20%29%7D%7B%5B%28%20C_%7B2%7DH_%7B5%7D%29_%7B3%7DN%5D%7D%20)
thus, [OH-] =(5.3 ^ 10-4) ^2 / 0.050
=0.00516 M
Thus, pOH = 2.30
pH = 14 - pOH = 11.7
Explanation:
To answer this question, we'll need to use the Ideal Gas Law:
p
V
=
n
R
T
,
where
p
is pressure,
V
is volume,
n
is the number of moles
R
is the gas constant, and
T
is temperature in Kelvin.
The question already gives us the values for
p
and
T
, because helium is at STP. This means that temperature is
273.15 K
and pressure is
1 atm
.
We also already know the gas constant. In our case, we'll use the value of
0.08206 L atm/K mol
since these units fit the units of our given values the best.
We can find the value for
n
by dividing the mass of helium gas by its molar mass:
n
=
number of moles
=
mass of sample
molar mass
=
6.00 g
4.00 g/mol
=
1.50 mol
Now, we can just plug all of these values in and solve for
V
:
p
V
=
n
R
T
V
=
n
R
T
p
=
1.50 mol
×
0.08206 L atm/K mol
×
273.15 K
1 atm
= 33.6 L
this is not the answer but it will help you
do by the formula it is on the answer
Answer:
I am pretty sure Danny Duncan told me 69
Explanation:
niice
