Energy is released and mass is reduced
Answer:
The correct answer is C: A radiation consisting of uncharged particles is emitted when alpha particles strike beryllium atoms.
Explanation:
Chadwick discovered neutron while experimenting with a gold foil. A stream of alpha particles produced from a polonium source was directed at beryllium target. It was noticed that some penetrating radiations were produced. These uncharged particles were called neutrons because on the charge detector these particles showed no deflection.
Their nuclear reaction is as follow.
₂He + ₄Be ----------> ₀n + ₆C
Chadwick noticed that neutrons cannot ionize gases and are highly penetrating particles.
Answer:
![r=25M^{-1}s^{-1}[A]^2](https://tex.z-dn.net/?f=r%3D25M%5E%7B-1%7Ds%5E%7B-1%7D%5BA%5D%5E2)
Explanation:
Hello there!
In this case, according to the given information for this chemical reaction, it is possible for us to set up the following general rate law and the ratio of the initial and the final (doubled concentration) condition:
![r=k[A]^n\\\\\frac{r_1}{r_2} =\frac{k[A]_1^n}{k[A]_2^n}](https://tex.z-dn.net/?f=r%3Dk%5BA%5D%5En%5C%5C%5C%5C%5Cfrac%7Br_1%7D%7Br_2%7D%20%3D%5Cfrac%7Bk%5BA%5D_1%5En%7D%7Bk%5BA%5D_2%5En%7D)
Next, we plug in the given concentrations of A, 0.2M and 0.4 M, the rates, 1.0 M/s and 4.0 M/s and cancel out the rate constants as they are the same, in order to obtain the following:
Which means this reaction is second-order with respect to A. Finally, we calculate the rate constant by using n, [A] and r, to obtain:
![k=\frac{r}{[A]^n} =\frac{1.0M/s}{(0.2M)^2}\\\\k=25M^{-1}s^{-1}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7Br%7D%7B%5BA%5D%5En%7D%20%3D%5Cfrac%7B1.0M%2Fs%7D%7B%280.2M%29%5E2%7D%5C%5C%5C%5Ck%3D25M%5E%7B-1%7Ds%5E%7B-1%7D)
Thus, the rate law turns out to be:
Regards!
Answer:
A liquid is a nearly incompressible fluid that conforms to the shape of its container but retains a (nearly) constant volume independent of pressure. As such, it is one of the four fundamental states of matter (the others being solid, gas, and plasma), and is the only state with a definite volume but no fixed shape.
Answer:
The answer to the question is
The height of the mercury fluid column remain the same.
Explanation:
The pressure, P in a column of fluid of height, h is given by
P = (Density p)×(height of fluid column h)×(gravity g)
Therefore, when the diameter is doubled we have
Density of the mercury in the tube with twice the diameter = (Mass of mercury)/(volume of mercury) where the volume of mercury = h×pi×(Diameter×2)^2/4 = h×pi×Diameter^2. Therefore the volume increases by a factor of 4 and therefore the mass increases by a factor of 4 which means that the density remains the same hence
P = p×h1×g = p×h2×g Therefore h1 = h2
The height of the fluid column remain the same
