Ok so this is what we know :
2KClO3 -> 2KCl + 3O2 (Always check if equation is balanced - in this case it is)
4.26moles
So we know that we have 4.26 moles of oxygen (O2). Now lets look at the ratio between KClO3 and O2.
We see that the ratio is 2:3 meaning that we need 2KClO3 in order to produce 3O2.
Therefore divide 4.26 by 3 and then multiply by 2.
4.26/3 = 1.42
1.42 * 2 = 2.84
Now we know that the molarity of KClO3 is 2.84 moles.
Multiply by R.M.M to find how many grams of KClO3 we have.
R.M.M of KClO3
K- 39
Cl- 35.5
3O- 3 * 16 -> 48
---------------------------
<span>122.5
</span>2.84 * 122.5 = 347.9 grams therefore the answer is (a)
348 grams needed of KClO3 to produce 4.26 moles of O2.
Hope this helps :).
Bromcresol green is the indicator that is blue in a solution that has a Ph of 5.6.
The ore contains 55.4% calcium phosphate (related to the mineral apatite) so the amount of Ca3(PO4)2 is 55.4%x=1000g so x=1000/0.554= 1.805kg. Now for the % of P in this amount of calcium phosphate, use all the masses of the elements in Ca3PO4= Ca=40.078 x 3= 120.23 and (PO4)2= (30.974+64)2=189.95 (NB oxygen is 16 mass x 4 =64) so the total mass is 310.2 and we have 61.95 of P (Pmass x 2) so 61.95/3102.= 0.19 or 19% P. So of the 1.805 x 0.19= 0.34kg of phosphorus.
Convert 57.6 L to dm3 and divide it by 24
Answer:
A) ∆Suniv >0, ∆G<0, T∆Suniv >0.
Explanation:
The connection between entropy and the spontaneity of a reaction is expressed by the <u>second law of thermodynamics</u><u>: The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process</u>.
Mathematically, we can express the second law of thermodynamics as follows:
For a spontaneous process: ΔSuniv = ΔSsys + ΔSsurr > 0
Therefore, the second law of thermodynamics tells us that a spontaneous reaction increases the entropy of the universe; that is, ΔSuniv > 0.
If we want spontaneity expressed only in terms of the properties of the system (ΔHsys and ΔSsys), we use the following equation:
-TΔSuniv = ΔHsys - TΔSsys < 0
That means that T∆Suniv >0.
This equation says that for a process carried out at constant pressure and temperature T, if the changes in enthalpy and entropy of the system are such that <u>ΔHsys - TΔSsys is less than zero, the process must be spontaneous.</u>
Finally, if the change in free energy is less than zero (ΔG<0), the reaction is spontaneous in the forward direction.
