Answer:
<h2>3.31 g/mL</h2>
Explanation:
The density of a substance can be found by using the formula
![density = \frac{mass}{volume} \\](https://tex.z-dn.net/?f=density%20%3D%20%20%5Cfrac%7Bmass%7D%7Bvolume%7D%20%5C%5C)
From the question
mass = 23.2 g
volume = final volume of water - initial volume of water
volume = 62 - 55 = 7 mL
We have
![density = \frac{23.2}{7} \\ = 3.314285](https://tex.z-dn.net/?f=density%20%3D%20%20%5Cfrac%7B23.2%7D%7B7%7D%20%20%5C%5C%20%20%3D%203.314285)
We have the final answer as
<h3>3.31 g/mL</h3>
Hope this helps you
Answer:
Rutherford's experiment, also known as
![\alpha - scattering \: experiment](https://tex.z-dn.net/?f=%20%5Calpha%20%20-%20scattering%20%5C%3A%20experiment)
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 quantity of heat required is 358.644 J.
Explanation:
Given: Specific heat capacity = ![4.18 J/g^{o}C](https://tex.z-dn.net/?f=4.18%20J%2Fg%5E%7Bo%7DC)
Mass = 1.50 g
![T_{1} = 26.5^{o}C](https://tex.z-dn.net/?f=T_%7B1%7D%20%3D%2026.5%5E%7Bo%7DC)
![T_{2} = 83.7^{o}C](https://tex.z-dn.net/?f=T_%7B2%7D%20%3D%2083.7%5E%7Bo%7DC)
Formula used to calculate heat energy is as follows.
![q = m \times C \times (T_{2} - T_{1})](https://tex.z-dn.net/?f=q%20%3D%20m%20%5Ctimes%20C%20%5Ctimes%20%28T_%7B2%7D%20-%20T_%7B1%7D%29)
where,
q = heat energy
m = mass
C = specific heat capacity
= initial temperature
= final temperature
Substitute the values into above formula as follows.
![q = m \times C \times (T_{2} - T_{1})\\= 1.50 \times 4.18 J/g^{o}C \times (83.7 - 26.5)^{o}C\\= 358.644 J](https://tex.z-dn.net/?f=q%20%3D%20m%20%5Ctimes%20C%20%5Ctimes%20%28T_%7B2%7D%20-%20T_%7B1%7D%29%5C%5C%3D%201.50%20%5Ctimes%204.18%20J%2Fg%5E%7Bo%7DC%20%5Ctimes%20%2883.7%20-%2026.5%29%5E%7Bo%7DC%5C%5C%3D%20358.644%20J)
Thus, we can conclude that quantity of heat required is 358.644 J.
I think a is your answer hope it's right