Data analysis is a process of inspecting, cleansing, transforming and modeling data with the goal of discovering useful information, informing conclusions and supporting decision-making
Answer: 0.52849 j /g °C
Explanation:
Given the following :
Mass of metal = 36g
Δ Temperature of metal = (28.4 - 99)°C = - 70.6°C
Mass of water = 70g
Δ in temperature of water = (28.4 - 24.0) = 4.4°C
Heat lost by metal = (heat gained by water + heat gained by calorimeter)
Quantity of heat(q) = mcΔT
Where; m = mass of object ; c = specific heat capacity of object
Heat lost by metal:
- (36 × c × - 70.6) = 2541.6c - - - - (1)
Heta gained by water and calorimeter :
(70 × 4.184 × 4.4) + (12.4 × 4.4) = 1288.672 + 54.56 = 1343.232 - - - - (2)
Equating (1) and (2)
2541.6c = 1343.232
c = 1343.232 / 2541.6
c = 0.52849 j /g °C
Answer:
Fe
Explanation:
<em>Ferrum</em><em> </em>[Iron] has the most stable nucleus because of <em>binding</em><em> </em><em>energy</em><em> </em><em>per</em><em> </em><em>nucleon</em><em>.</em><em> </em>Although Uranium<em> </em>is<em> </em>another possibility, in this case, it is more radioactive than Iron. It disintegrates very swiftly that all is done so, just to achieve stability.
I hope this helps you out alot, and as always, I am joyous to assist anyone at any time.
Explanation:
because translucent shades lets the light through easily (gentle diffusion)
<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