It results in a chemical reaction called neutraluization.
<u>Answer:</u> The
for the reaction is -297 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 given chemical reaction follows:
![\Delta H^o_{rxn}=?](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%3F)
The intermediate balanced chemical reaction are:
(1)
![\Delta H_1=-196kJ](https://tex.z-dn.net/?f=%5CDelta%20H_1%3D-196kJ)
(2)
![\Delta H_2=-790kJ](https://tex.z-dn.net/?f=%5CDelta%20H_2%3D-790kJ)
The expression for enthalpy of the reaction follows:
![\Delta H^o_{rxn}=\frac{[1\times (-\Delta H_1)]+[1\times \Delta H_2]}{2}](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5Cfrac%7B%5B1%5Ctimes%20%28-%5CDelta%20H_1%29%5D%2B%5B1%5Ctimes%20%5CDelta%20H_2%5D%7D%7B2%7D)
Putting values in above equation, we get:
![\Delta H^o_{rxn}=\frac{[(1\times -(-196))+(1\times (-790))}{2}=-297kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5Cfrac%7B%5B%281%5Ctimes%20-%28-196%29%29%2B%281%5Ctimes%20%28-790%29%29%7D%7B2%7D%3D-297kJ)
Hence, the
for the reaction is -297 kJ.
Answer:
D-Molten rock cools and turns solid
pH value of a 0. 011 m naf solution is 8.57.
Solution
This problem uses the relationship between Kb and the dissociation constants which is expressed as Kw = KaKb. Calculations are as follows:
Kb = KaKb
1.00 x 10^-14 = 7.2 x 10^-4(x)
x = 1.39 x 10^-11
We now need to calculate the [OH¯] using the Kb expression:
1.39 x 10^-11 = x^2 / (0.30 - x)
The denominator can be neglected. Thus, x is 3.73 x 10^-6.
pOH = -log 3.73 x 10^-6 = 5.43
pH = 14-5.43 = 8.57
To learn more ph of naf solution refer here:
brainly.com/question/1411794
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Answer:
0.0933 moles/Litre
Explanation:
We assume that the number of moles of N- used is equal to the number of moles of Nitrogen containing compounds that are generated due to the fact that the nitrogen containing compound that are produced contain only one nitrogen in each atom. As such, finding the amount of nitrogen used up explains the amount of compound formed. This can be expressed as follows:
Energy cost = ![\frac{20ev}{N- atom}](https://tex.z-dn.net/?f=%5Cfrac%7B20ev%7D%7BN-%20atom%7D)
Given that:
Energy = 100 W for 60 minutes
100 W = 100 J/s
= 100 J/s × (60 × 60) seconds
= 3.6 × 10⁵ J
Let now convert 3.6 × 10⁵ J to eV; we have:
= ( 3.6 × 10⁵ × 6.242 × 10¹⁸ )eV
= 2.247 × 10²⁴ eV
So, number of N-atom used up to form compounds will now be:
= 2.247 × 10²⁴ eV × ![\frac{1}{20eV} N-atom](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B20eV%7D%20N-atom)
= 1.123 × 10²³ N-atom
To moles; we have:
= ![\frac {1.123*10^{23}} {6.023*10^{23}}](https://tex.z-dn.net/?f=%5Cfrac%20%7B1.123%2A10%5E%7B23%7D%7D%20%7B6.023%2A10%5E%7B23%7D%7D)
= 0.186 moles
However, we are expected to leave our answer in concentration (i.e in moles/L)
since we are given 2L
So; 0.186 moles ⇒ ![\frac{0.186 moles}{2 Litre}](https://tex.z-dn.net/?f=%5Cfrac%7B0.186%20moles%7D%7B2%20Litre%7D)
= 0.0933 moles/Litre