Answer:
Acceleration due to gravity is reduced to half its value on the earth's surface at an altitude of 2.65×106 m
Answer:
b
Explanation:
A pure substance has a constant composition and cannot be separated into simpler substances by physical means. There are two types of pure substances: elements and compounds.
Answer:
1. Changing Beam Material
2. Corrugation
3. Changing Beam form
4. Steel Reinforcing Bars
Explanation:
Changing Beam Material
Some materials are stronger when used in beams than others. Beams made of steel for instance are stronger than beams made of wood. Therefore changing material can improve the strength of the beam. It is quite important to take into account the weights of the material though as different structures have different requirements.
Corrugation.
You can fold the beam into triangular shapes to increase strength. If you look at roofs you will notice that they are folded and this increased their strength. The same logic can be applied to beams.
Changing Beam Form
Another way to make Beams stronger is to change their form or rather their shape. Straight beams are not as strong as I-beams for instance. I-beams look like the capital letter I with the lines at both ends. I-beams are usually used in construction which shows that they are quite strong.
Steel Reinforcing Bars
When placed in concrete beams, Steel Reinforcing Bars which are also called Rebar can help strengthen a beam by helping it withstand the forces of tension. A concrete beam with Rebar inside it is known as Reinforced Concrete.
<u>Answer:</u> The standard free energy change of formation of 
is 92.094 kJ/mol
<u>Explanation:</u>
We are given:
Relation between standard Gibbs free energy and equilibrium constant follows:
where,
= standard Gibbs free energy = ?
R = Gas constant = 
T = temperature = ![25^oC=[273+25]K=298K](https://tex.z-dn.net/?f=25%5EoC%3D%5B273%2B25%5DK%3D298K)
K = equilibrium constant or solubility product = 
Putting values in above equation, we get:
For the given chemical equation:
The equation used to calculate Gibbs free change is of a reaction is:
![\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28reactant%29%7D%5D)
The equation for the Gibbs free energy change of the above reaction is:
![\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(Ag^+(aq.))})+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times \Delta G^o_f_{(Ag_2S(s))})]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28Ag%5E%2B%28aq.%29%29%7D%29%2B%281%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28S%5E%7B2-%7D%28aq.%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28Ag_2S%28s%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![285.794=[(2\times 77.1)+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times (-39.5))]\\\\\Delta G^o_f_{(S^{2-}(aq.))=92.094J/mol](https://tex.z-dn.net/?f=285.794%3D%5B%282%5Ctimes%2077.1%29%2B%281%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28S%5E%7B2-%7D%28aq.%29%29%7D%29%5D-%5B%281%5Ctimes%20%28-39.5%29%29%5D%5C%5C%5C%5C%5CDelta%20G%5Eo_f_%7B%28S%5E%7B2-%7D%28aq.%29%29%3D92.094J%2Fmol)
Hence, the standard free energy change of formation of is 92.094 kJ/mol
Answer:
Answer D => E°(Mg°/Cu⁺²) = 0.34 + 2.37 = 2.71v
Explanation:
(Oxidation) => Mg°(s) => Mg⁺²(aq) + 2e⁻ E°(Mg°/Mg⁺²) = -2.37 v
(Reduction) => Cu⁺²(aq) + 2e⁻ => Cu°(s) E°(Cu⁺²/Cu°) = +0.34 v
________________________________________________
Net Rxn => Mg°(s) + Cu⁺²(aq) => Mg⁺²(aq) + Cu°(s)
Std Cell Potential (25°C/1Atm) = E°(Redn) = E°(Oxidn) = +0.34v - (-2.37v)
= 0.34v + 2.37v = 2.72v
