When atoms of beryllium-9 are bombarded with alpha particles, neutrons are produced, the new isotope will be C-6.
In research, the physicist James Chadwick attacked Beryllium using alpha particles produced by the nuclear reactions of polonium naturally. High radiation penetration through some kind of lead shield was seen, which the then-current particle theories were unable to account for.
There was indeed the production of a carbon-12 nucleus as well as the production of a neutron whenever one blasted beryllium with alpha particles.
The reaction of alpha decay can be written as:
→ ![C^{12} _{6} + n^{1} _{0}](https://tex.z-dn.net/?f=C%5E%7B12%7D%20_%7B6%7D%20%2B%20n%5E%7B1%7D%20_%7B0%7D)
Therefore, when atoms of beryllium-9 are bombarded with alpha particles, neutrons are produced, the new isotope will be C-6.
To know more about alpha particles
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Answer:
Explanation:
The equation of above line , y = 0.0005x+ 0.458
This can be compared with y = mx+c
Hence slope, m = 0.0005 and Y-intercept, c = 0.458
Or it can be plotted manually where straight line has to be drawn touching maximum number of data points. After drawing a straight linear line, we need to take any two points from the straight line and slope is calculated
Slope,
![m = \frac{y_2-y_1}{x_2-x_1}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7By_2-y_1%7D%7Bx_2-x_1%7D)
and y -intercept is calculated using extraplotting backwards such that it touches the Y-axis. the point where straight line touches Y-axis is Y-intercept (c).
Plot the average cell potentials E (y-axis) vs T (x-axis). image attached
Answer:
1.19 g
Explanation:
Given that:
Molecular weight of diene
136 g/mol
Molecular weight of Maleic Anhydride
96 g/mol
Mass of crude diene = 2.5 g
Percent of Composition for Peak A = 66%
Percent of Composition for Peak B = 22.66%
Percent of Composition for Peak C = 11.33%
Let us determine the amount of diene in the sample;
So, using
to represent the amount of diene; we have :
= ![m*P_a](https://tex.z-dn.net/?f=m%2AP_a)
= ![2.5 *\frac{66}{100}](https://tex.z-dn.net/?f=2.5%20%2A%5Cfrac%7B66%7D%7B100%7D)
= 1.65 g
Using the limiting reagent equation to determine the amount of Maleic Anhydride ![A_{ma}](https://tex.z-dn.net/?f=A_%7Bma%7D)
=
![* \frac{1mole diene}{M.W_d} *\frac{M.W_{m.a}}{1 mole of maleic anhydride}](https://tex.z-dn.net/?f=%2A%20%5Cfrac%7B1mole%20diene%7D%7BM.W_d%7D%20%2A%5Cfrac%7BM.W_%7Bm.a%7D%7D%7B1%20mole%20of%20maleic%20anhydride%7D)
= ![1.65 * \frac{1 mole}{136}*\frac{98}{1 mole}](https://tex.z-dn.net/?f=1.65%20%2A%20%5Cfrac%7B1%20mole%7D%7B136%7D%2A%5Cfrac%7B98%7D%7B1%20mole%7D)
= 1.19 g
The grams of Maleic Anhydride used to react with 2.5 g sample of crude diene = 1.19 g
It does not create another substance or new substance.
Hope this helps