Hey it's me again haha!
the answer to your question is:
answer: so f=20.1
a=8.2
m=?
f=ma
20.1 = 8.2*m in
20.1/8.2=m
so the answer is around 2.45 something!
also i'm sorry this question had be stumbled!
Answer : The equilibrium concentration of
at
is,
.
Solution : Given,
Equilibrium constant, 
Initial concentration of
= 0.260 m
Let, the 'x' mol/L of
are formed and at same time 'x' mol/L of
are also formed.
The equilibrium reaction is,

Initially 0.260 m 0 0
At equilibrium (0.260 - x) x x
The expression for equilibrium constant for a given reaction is,
![K_c=\frac{[H_3O^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BC_2H_3O_2%5E-%5D%7D%7B%5BHC_2H_3O_2%5D%7D)
Now put all the given values in this expression, we get

By rearranging the terms, we get the value of 'x'.

Therefore, the equilibrium concentration of
at
is,
.
Answer:
use secondary data. the normal method to use
When liquid state is being converted to solid state, atoms come closer to each other and their speed of movement also decreases. Instead of moving about in the substance, they are only able to vibrate about their positions in solid. Hence, this change leads to loss of energy. Similarly, atoms loose energy when state changes from gas to liquid.
2075g of XeF₆ is present in 8.46 moles of XeF₆
<u>Explanation:</u>
Given:
XeF₆ is xenon hexafluoride
number of moles of XeF₆ is 8.46
Mass of XeF₆ present, m = ?
We know,
Molecular weight of XeF₆, M is 245.28 g/mol
and
number of moles, n = mass of XeF₆ present / molecular weight of XeF6

Therefore, 2075g of XeF₆ is present in 8.46 moles of XeF₆