Answer:
39.6 g
Explanation:
The equation of the reaction is;
2Mg(s) + O2(g) --------> 2MgO(s)
To obtain the limiting reactant;
Number of moles in 26.4 g of Mg = 26.4g/24 g/mol = 1.1 moles
If 2 moles of Mg yields 2 moles of MgO
1.1 moles of Mg yields 1.1 * 2/2 = 1.1 moles of MgO
Number of moles in 26.4 g of O2 = 26.4 g/32g/mol = 0.825 moles
If 1 mole of O2 yields 2 moles of MgO
0.825 moles of O2 yields 0.825 moles * 2/1 = 1.65 moles of MgO
Hence Mg is the limiting reactant.
Theoretical yield of MgO = 1.1 moles of MgO * 40 g/mol = 44 g
Percent yield = 90%
Percent yield = actual yield/theoretical yield * 100
Actual yield = Percent yield * theoretical yield/100
Actual yield = 90 * 44/100
Actual yield = 39.6 g
Answer:
120.575 kJ is the activation energy for the souring process.
Explanation:
The formula for an activation energy is given as:
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= rate constant at 
= 
= rate constant at 
= 
= activation energy for the reaction = ?
R = gas constant = 8.314 J/mole.K
= initial temperature = 
= final temperature = 
Now put all the given values in this formula, we get:l
![\log (\frac{k}{40k})=\frac{Ea}{2.303\times 8.314 J/mol K}[\frac{1}{298K}-\frac{1}{277 K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7Bk%7D%7B40k%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%208.314%20J%2Fmol%20K%7D%5B%5Cfrac%7B1%7D%7B298K%7D-%5Cfrac%7B1%7D%7B277%20K%7D%5D)
120.575 kJ is the activation energy for the souring process.
Answer:
by wearing of rocks
Explanation:
earths gravity your welcome
Since
potassium and phosphate is what we are to find for and they are both found in
the potassium phosphate solution, therefore we solve for this one first on the
basis of the phosphate.
The formula
for finding the volume given the concentration and number of moles is:
Volume =
number of moles / concentration in Molarity
Volume
potassium phosphate required = 30 mmol phosphate / (3 mmol / mL)
<u>Volume
potassium phosphate required = 10 mL</u>
This would
also contain potassium in amounts of:
Amount of
potassium in potassium phosphate = 10 mL (4.4 meg / mL)
Amount of
potassium in potassium phosphate = 44 meg
Therefore
the potassium chloride required is:
Volume of
potassium chloride = (80 meg – 44 meg) / (2 meg / mL)
<span><u>Volume of
potassium chloride = 72 mL</u></span>
