Explanation:
Rutherford conducted an experiment in which he took a thin gold particle film on which he passes alpha- particles. He noticed that:
- Most of the alpha particles get through the film and can be detected by the detector.
- Around small portion of the alpha particle deflected at small angles.
- A very very few alpha particle (approximately 1 out of 1 million alpha particles) just retraced their path which means come back from the center.
He concluded that:
<u>Most of the space of the atom is empty and in the center of the atom , there is solid mass which is the cause of the alpha particles to come back. He gave the term nucleus to this solid mass.</u>
Answer:
-3.7771 × 10² kJ/mol
Explanation:
Let's consider the following equation.
3 Mg(s) + 2 Al³⁺(aq) ⇌ 3 Mg²⁺(aq) + 2 Al(s)
We can calculate the standard Gibbs free energy (ΔG°) using the following expression.
ΔG° = ∑np . ΔG°f(p) - ∑nr . ΔG°f(r)
where,
n: moles
ΔG°f(): standard Gibbs free energy of formation
p: products
r: reactants
ΔG° = 3 mol × ΔG°f(Mg²⁺(aq)) + 2 mol × ΔG°f(Al(s)) - 3 mol × ΔG°f(Mg(s)) - 2 mol × ΔG°f(Al³⁺(aq))
ΔG° = 3 mol × (-456.35 kJ/mol) + 2 mol × 0 kJ/mol - 3 mol × 0 kJ/mol - 2 mol × (-495.67 kJ/mol)
ΔG° = -377.71 kJ = -3.7771 × 10² kJ
This is the standard Gibbs free energy per mole of reaction.
Answer:
b. Magnetism (sorry im very late)
Explanation:
Intensive properties do not depend on size, no matter what it doesn't. For example, magnetism, density, melting and boiling points, and color. All of those support intensive property.
Answer:
526.85K
Explanation:
Based on Charles's law, the volume of a gas is directly proportional to absolute temperature. The formrula is:
V₁ / T₁ = V₂ / T₂
<em>Where 1 represents the initial state and 2 the final state of the gas</em>
Using the values of the problem:
V₁ = 10.0L
T₁ = 127°C + 273.15K = 400.15K
V₂ = 20.0L
Thus, replacing in the formula:
10.0L / 400.15K = 20.0L / T₂
T₂ = 800K
In Celsius:
800K - 273.15 =<em> 526.85K</em>
<em />
<u>Answer:</u> The balanced chemical equation is written below and 
for the reaction is -160.6 J/K
<u>Explanation:</u>
When calcium hydroxide reacts with sulfur dioxide, it leads to the formation of calcium sulfate and water molecule.
The chemical equation for the reaction of calcium hydroxide and sulfur dioxide follows:
To calculate the entropy change of the reaction, we use the equation:
![\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{products}]-\sum [n\times \Delta S^o_{reactants}]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20S%5Eo_%7Bproducts%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20S%5Eo_%7Breactants%7D%5D)
For the given reaction:
![\Delta S^o_{rxn}=[(1\times \Delta S^o_{CaSO_3(s)})+(1\times \Delta S^o_{H_2O(l)})]-[(1\times \Delta S^o_{Ca(OH)_2(s)})+(1\times \Delta S^o_{SO_2(g)})]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20S%5Eo_%7BCaSO_3%28s%29%7D%29%2B%281%5Ctimes%20%5CDelta%20S%5Eo_%7BH_2O%28l%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20S%5Eo_%7BCa%28OH%29_2%28s%29%7D%29%2B%281%5Ctimes%20%5CDelta%20S%5Eo_%7BSO_2%28g%29%7D%29%5D)
Taking the standard entropy change values:
Putting values in above equation, we get:
![\Delta S^o_{rxn}=[(1\times (101.4))+(1\times (69.9))]-[(1\times (83.4))+(1\times (248.5))]\\\\\Delta S^o_{rxn}=-160.6J/K](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%28101.4%29%29%2B%281%5Ctimes%20%2869.9%29%29%5D-%5B%281%5Ctimes%20%2883.4%29%29%2B%281%5Ctimes%20%28248.5%29%29%5D%5C%5C%5C%5C%5CDelta%20S%5Eo_%7Brxn%7D%3D-160.6J%2FK)
Hence, the balanced chemical equation is written above and for the reaction is -160.6 J/K
