<u>Answer:</u> The standard heat for the given reaction is -138.82 kJ
<u>Explanation:</u>
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.
The equation used to calculate enthalpy change is of a reaction is:
![\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f_%7B%28reactant%29%7D%5D)
For the given chemical reaction:
The equation for the enthalpy change of the above reaction is:
![\Delta H_{rxn}=[(3\times \Delta H_f_{(CH_4(g))})+(1\times \Delta H_f_{(CO_2(g))})+(4\times \Delta H_f_{(NH_3(g))})]-[(4\times \Delta H_f_{(CH_3NH_2(g))})+(2\times \Delta H_f_{(H_2O(l))})]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%283%5Ctimes%20%5CDelta%20H_f_%7B%28CH_4%28g%29%29%7D%29%2B%281%5Ctimes%20%5CDelta%20H_f_%7B%28CO_2%28g%29%29%7D%29%2B%284%5Ctimes%20%5CDelta%20H_f_%7B%28NH_3%28g%29%29%7D%29%5D-%5B%284%5Ctimes%20%5CDelta%20H_f_%7B%28CH_3NH_2%28g%29%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H_f_%7B%28H_2O%28l%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta H_{rxn}=[(3\times (-74.8))+(1\times (-393.5))+(4\times (-46.1))]-[(4\times (-22.97))+(2\times (-285.8))]\\\\\Delta H_{rxn}=-138.82kJ](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%283%5Ctimes%20%28-74.8%29%29%2B%281%5Ctimes%20%28-393.5%29%29%2B%284%5Ctimes%20%28-46.1%29%29%5D-%5B%284%5Ctimes%20%28-22.97%29%29%2B%282%5Ctimes%20%28-285.8%29%29%5D%5C%5C%5C%5C%5CDelta%20H_%7Brxn%7D%3D-138.82kJ)
Hence, the standard heat for the given reaction is -138.82 kJ
Answer:
0.9715 Fraction of Pu-239 will be remain after 1000 years.
Explanation:
Where:
= decay constant
=concentration left after time t
= Half life of the sample
Half life of Pu-239 = 
[
![\lambda =\frac{0.693}{24,000 y}=2.8875\times 10^{-5} y^{-1]](https://tex.z-dn.net/?f=%5Clambda%20%3D%5Cfrac%7B0.693%7D%7B24%2C000%20y%7D%3D2.8875%5Ctimes%2010%5E%7B-5%7D%20y%5E%7B-1%5D)
Let us say amount present of Pu-239 today = 
A = ?
![A=x\times e^{-2.8875\times 10^{-5} y^{-1]\times 1000 y}](https://tex.z-dn.net/?f=A%3Dx%5Ctimes%20e%5E%7B-2.8875%5Ctimes%2010%5E%7B-5%7D%20y%5E%7B-1%5D%5Ctimes%201000%20y%7D)
0.9715 Fraction of Pu-239 will be remain after 1000 years.
Some of the reactants or the products are in the gaseous phase.
The hydrogen bonding in H₂O is stronger than that of HF
Explanation:
Hydrogen bonds are special dipole-dipole attraction in which electrostatic attraction is established between hydrogen atom of one molecule and the electronegative atom of a neighboring molecule.
- The strength of hydrogen bonds depends on the how electronegative an atom is.
- Electronegativity refers to the tendency of an atom to gain electrons.
- The higher the value, the higher the tendency.
- This why oxygen with a higher electronegativity will form a stronger hydrogen bond with hydrogen compared to fluorine.
Learn more:
hydrogen bond brainly.com/question/12408823
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