Answer:
7.37 mL of KOH
Explanation:
So here we have the following chemical formula ( already balanced ), as HNO3 reacts with KOH to form the products KNO3 and H2O. As you can tell, this is a double replacement reaction,
HNO3 + KOH → KNO3 + H2O
Step 1 : The moles of HNO3 here can be calculated through the given molar mass ( 0.140 M HNO3 ) and the mL of this nitric acid. Of course the molar mass is given by mol / L, so we would have to convert mL to L.
Mol of NHO3 = 0.140 M 
30 / 1000 L = 0.140 M 0.03 L = .0042 mol
Step 2 : We can now convert the moles of HNO3 to moles of KOH through dimensional analysis,
0.0042 mol HNO2 ( 1 mol KOH / 1 mol HNO2 ) = 0.0042 mol KOH
From the formula we can see that there is 1 mole of KOH present per 1 moles of HNO2, in a 1 : 1 ratio. As expected the number of moles of each should be the same,
Step 3 : Now we can calculate the volume of KOH knowing it's moles, and molar mass ( 0.570 M ).
Volume of KOH = 0.0042 mol ( 1 L / 0.570 mol ) ( 1000 mL / 1 L ) = 7.37 mL of KOH
<u>Answer:</u> The 
for the reaction is -1406.8 kJ.
<u>Explanation:</u>
Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.
According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.
The chemical reaction for the formation reaction of 
is:
The intermediate balanced chemical reaction are:
(1) 
( × 6)
(2) 
( × 3)
(3) 
( × 2)
(4) 
The expression for enthalpy of formation of is,
![\Delta H^o_{formation}=[6\times \Delta H_1]+[3\times \Delta H_2]+[2\times \Delta H_3]+[1\times \Delta H_4]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Bformation%7D%3D%5B6%5Ctimes%20%5CDelta%20H_1%5D%2B%5B3%5Ctimes%20%5CDelta%20H_2%5D%2B%5B2%5Ctimes%20%5CDelta%20H_3%5D%2B%5B1%5Ctimes%20%5CDelta%20H_4%5D)
Putting values in above equation, we get:
![\Delta H^o_{formation}=[(-74.8\times 6)+(-185\times 3)+(323\times 2)+(-1049\times 1)]=-1406.8kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Bformation%7D%3D%5B%28-74.8%5Ctimes%206%29%2B%28-185%5Ctimes%203%29%2B%28323%5Ctimes%202%29%2B%28-1049%5Ctimes%201%29%5D%3D-1406.8kJ)
Hence, the for the reaction is -1406.8 kJ.
Hey there:
a) atomic mass:
Carbon =<span>12.0107 g/mol
</span>Hydrogen = <span>1.00794 g/mol
Oxygen = </span><span>15.9994 g/mol
</span>
Therefore:
C12H22O11 =
12 * 12.0107 + 1 * 1.00794 + 16 * 15.9994 => <span>342.29648 g/mol
__________________________________________________________
b) number of moles:
n = m / mm
n = 3.115 / </span><span>342.29648
n = 0.0091 moles
________________________________________________
hope this helps!</span>
