Answer:
It could lead to Fatigue
Low energy.
Headaches.
Aches, pains, and tense muscles.
Chest pain and rapid heartbeat.
Explanation:
Answer: -105 kJ
Explanation:-
The balanced chemical reaction is,

The expression for enthalpy change is,
![\Delta H=\sum [n\times B.E(reactant)]-\sum [n\times B.E(product)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Csum%20%5Bn%5Ctimes%20B.E%28reactant%29%5D-%5Csum%20%5Bn%5Ctimes%20B.E%28product%29%5D)
![\Delta H=[(n_{N_2}\times B.E_{N_2})+(n_{H_2}\times B.E_{H_2}) ]-[(n_{NH_3}\times B.E_{NH_3})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BN_2%7D%5Ctimes%20B.E_%7BN_2%7D%29%2B%28n_%7BH_2%7D%5Ctimes%20B.E_%7BH_2%7D%29%20%5D-%5B%28n_%7BNH_3%7D%5Ctimes%20B.E_%7BNH_3%7D%29%5D)
![\Delta H=[(n_{N_2}\times B.E_{N\equiv N})+(n_{H_2}\times B.E_{H-H}) ]-[(n_{NH_3}\times 3\times B.E_{N-H})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BN_2%7D%5Ctimes%20B.E_%7BN%5Cequiv%20N%7D%29%2B%28n_%7BH_2%7D%5Ctimes%20B.E_%7BH-H%7D%29%20%5D-%5B%28n_%7BNH_3%7D%5Ctimes%203%5Ctimes%20B.E_%7BN-H%7D%29%5D)
where,
n = number of moles
Now put all the given values in this expression, we get
![\Delta H=[(1\times 945)+(3\times 432)]-[(2\times 3\times 391)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%281%5Ctimes%20945%29%2B%283%5Ctimes%20432%29%5D-%5B%282%5Ctimes%203%5Ctimes%20391%29%5D)

Therefore, the enthalpy change for this reaction is, -105 kJ
Answer:
( °F − 32) × 5/9 = °C
Explanation:
Also there is a mental calculation to convert from Fahrenheit to Celsius. The ratio 5/9 is approximately equal 0.55555….
Subtract 32º to adapt the equivalent in the Fahrenheit scale.
Divide the degrees Celsius by 2 (multiply by 0.5).
Take 1/10 of this number (0.5 * 1/10 = 0.05) and add it to the number obtained previously.
Example: Convert 98.6º F to Centigrade.
98.6 - 32 = 66.6
66.6 * 1/2 = 33.3
33.3 * 1/10 = 3.3
33.3 + 3.3 = 36.6 which is an approximation in degrees Centigrade
The correct answer is: [A]: "<span>points with the same elevation" .
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Option (a) is correct.
A reducing agent is the one which loses electrons to other substance.
Here, Zn has oxidation number 0 in the L.H.S of the equation, but on R.H.S its oxidation number is +2 i.e. it Zn has donated two of its electrons to

.
Hence, Zn is the reducing agent here.