Answer:
Rutherford's experiment, also known as
supports the existence of neutrons and the nucleus.
Explanation:
In the above diagram, Rutherford was trying to explain his contributions using thin foils of gold and other metals as targets for alpha particles from a radioactive source.
He observed that the majority of particles penetrated the foil either undeflected or with only a slight deflection. But, every now and then an alpha particle was scattered(or deflected) at a large angle..
According to Rutherford, most of the atoms must be empty space. This explains why the majority of alpha particles passed through through the gold foil with little or no deflection. The atoms positive charges, Rutherford proposed are all concentrated in the Nucleus, <em>which</em><em> </em><em>is</em><em> </em><em>a</em><em> </em><em>dense</em><em> </em><em>central</em><em> </em><em>core</em><em> </em><em>withi</em><em>n</em><em> </em><em>the</em><em> </em><em>atom</em><em>. </em>
Whenever an alpha particle came close to a nucleus in the scattering experiment, it experienced a large repulsive force and therefore a large deflection. Moreover, an alpha particle coming towards a nucleus would be completely repelled and its direction would be reversed. The positively charged particles in the Nucleus are called Protons.
I <em>hope</em><em> </em><em>you</em><em> </em><em>find</em><em> </em><em>this</em><em> </em><em>useful</em><em>.</em><em>.</em><em>. </em><em>Have</em><em> </em><em>a</em><em> </em><em>lovely</em><em> </em><em>day</em><em>. </em>
Answer:
The enthalpy change per mole of Mg is (ΔH) = 460 kj mol⁻¹
Explanation:
the total volume of the solution is
100 ml, its mass is (100 ml)(1.0 g ml⁻¹) = 100 g (Density of water 1 g ml⁻¹)
The temperature change is ΔT = 11.1 ∘C
Heat of reaction = Cs × m × ∆T
= (4.18 Jg⁻¹ ∘C⁻¹)(100 g)(11.1 ∘C)
= 4639.8 j = 4.6 kJ
Because the process occurs at constant pressure, ΔH = qP = 4.6 kJ
To express the enthalpy change on a molar basis
Mole of Mg = = 0.01 mol
Thus, the enthalpy change per mole of Mg is ΔH = = 460 kj mol⁻¹
The overall reaction is:
Br₂(g) + 2 NO(g) ↔ 2 NOBr(g)
rate law = k [Br₂][NO]²
The first step of the overall reaction is:
NO(g) + Br₂(g) K₁⇄⇄K-1 NOBr₂(g)
rate law 1 = k₁ [Br₂][NO] or
rate law 2 = k-1 [NOBr₂]
The second step of the overall reaction is:
NOBr₂(g) + NO(g) →K₂→ 2 NOBr
rate law 3 = k₂[NOBr₂][NO]
So, rate law of overall reaction can be obtained as follows:
(rate law 1)*(rate law 3) / (rate law 2)
= [(k₁[Br₂][NO])* (K₂[NOBr₂][NO])] / k₋₁[NOBr₂]
= [k₁k₂/k₋₁][NO]²[Br₂]
So the correct answer is:
[k₁k₂/k₋₁][NO]² [Br₂]
Answer:
44.01 g/mol
Explanation:
Add each elements atomic mass. For oxygen you will do that twice because their is two oxygens.
- Hope that helps! Please let me know if you need further explanation.
Answer: 1 mole of strontium, 2 moles of hydrogen, 2 moles of carbon, 6 moles of O
Explanation: i think i’m not sure