Explanation:
The given reaction will be as follows.
So, equilibrium constant for this equation will be as follows.
![K_{c} = \frac{[CH_{3}OH]}{[CO][H_{2}]^{2}}](https://tex.z-dn.net/?f=K_%7Bc%7D%20%3D%20%5Cfrac%7B%5BCH_%7B3%7DOH%5D%7D%7B%5BCO%5D%5BH_%7B2%7D%5D%5E%7B2%7D%7D)
As it is given that concentration of all the species is 2.4. Therefore, calculate the value of equilibrium constant as follows.
= 
= 0.173
Thus, we can conclude that equilibrium constant for the given reaction is 0.173.
Answer:
those have symbols for their Latin or Greek name
Explanation:
hope it helps
You have to calculate the oxidation estates of the atoms in each compound.
I will start with K2Cr2O7 because I believe that Cr is the best candidate to reduce its oxidation number in 3 units.
In K2Cr2O7:
- K has oxidation state of 1+, then K2 has a charge of 2* (1+) = 2+.
- O has oxidation state of 2*, then O7 has a charge of 7* (2-) = 14-.
That makes that Cr2 has charge of 14 - 2 = +12, so each Cr has +12/2 = +6 oxidation state.
In Cr2O3:
- O has oxidation state of 2-, then O3 has charge 3 * (2-) = - 6
- Then, Cr2 has charge 6+, and each Cr has charge 6+ / 2 = 3+.
So, we have seen that Cr reduced its oxidation state in 3 units, from 6+ to 3+.
Answer: Cr has a change in oxidation number of - 3.
213034 torr is the osmotic pressure.
Explanation:
osmotic pressure is calculated by the formula:
osmotic pressure= iCrT
where i= no. of solute
c= concentration in mol/litre
R= Universal Gas constant
T = temp
It is given that solution is 3% which is 3gms in 100 ml.
let us calculate the concentration in moles/litre
3gm/100ml*1000ml/1L*1mol NaCl/55.84g NaCl
= 5.372 gm/litre
Putting the values in the formula, Temp in Kelvin 318.5K
osmotic pressure= 2*5.372*0.083 * 318.5 Gas constant 0.083
= 284.023 bar or 213018 torr. c= 5.372 moles/L
i=2 for NaCl
Chemical change creates a new substance
