increases my factor of 10
Answer:
1.33 Å
Explanation:
Given that the edge length , a of the KCl which forms the FCC lattice = 6.28 Å
Also,
For the FCC lattice in which the anion-cation contact along the cell edge , the ratio of the radius of the cation to that of anion is 0.731.
Thus,
.................1
Also, the sum of the radius of the cation and the anion in FCC is equal to half of the edge length.
Thus,
...................2
Given that:
To find,
Using 1 and 2 , we get:
<u>Size of the potassium ion = 1.33 Å</u>
<span>I did some investigation and summarized the process and made a clearer explanation so those who are confused can imagine the process better :) A scientific theory attempts to explain and describe why things happen. Hypotheses are formed and experiments are done to validate or toss the hypothesis based on the data collected. The Atomic Theory has gone through lots of refining as a scientific theory. For instance, William Crookes conduced an experiment with cathode ray tubes powered by electricity that glowed when powered. Crookes placed an object in between the positive and negative electrode and concluded that the shadow made on the positive side was small particles of matter traveling from the negative side. But more evidence was needed so, later on, J.J. Thomson continued Crookes experiment. He tested what would happen if a negative or positive charged rod was placed along the ray tubes and if it would differ if a different element was used as the negative electrode. Thomson found out that the beam had negatively charged particles and that even if the negative electrode is substituted, the glow is still present, meaning that all elements also had the small negative particles. These particles(now known as electrons) were smaller than the atom and were added to the model of the atom dispersed throughout the neutrally charged atom inside its positive sphere. Now came along Rutherford hoping to support Thomsons model by firing positively charged particles at a thin gold foil thinking it would go straight through the foil, but instead it evenly distributed as they went through the foil, concluding that atoms have a small, dense nucleus(containing positive protons and most of the mass of the atom) that deflected the particles passing through. This was a drastic change in the model now knowing that 1 proton has 2000 times the mass of an electron, but its positive charge cancels the negative electron. After WW1, Chadwick and others were seeing that sometimes the mass of the atom was greater than the mass of the protons and the number of protons was less than the mass of the atom. So it was thought that there were extra electrons and protons adding mass in the nucleus but cancelling their charges, but Rutherford proposed a particle with mass but no charge and called it a neutron; made of paired protons and electrons. But scientists kept studying atoms since there was no evidence of the neutron. Chadwick repeated these experiments though, in hopes to find the neutron and succeeded in 1932, finding it in the nucleus with a close mass to the proton. Thanks to these experiments for refining a scientific theory, we now have a clearer model of the atom.</span>
8 moles of water on the right side.
An oxidation-reduction or redox reaction is a reaction that involves the transfer of electrons between chemical items (the atoms, ions, or molecules involved in the reaction).
Redox reactions: the burning of fuels, the corrosion of metals, and even the processes of photosynthesis and cellular respiration involve oxidation and reduction.
Step 1:
MnO4- ----> Mn2+
2Cl- ------> Cl2
Step 2:
MnO4- --> Mn2+ + 4H2O
2Cl- -----> Cl2
Step 3:
8H+ + MnO4- ------> Mn2+ + 4H2O
2Cl- ----->Cl2
Step 4:
8H+ + MnO4- +5e- ------>Mn2+ + 4H2O
2Cl- ----> Cl2+ 2e-
Step 5:
16 H+ +2 MnO4- +10Cl- ----->2 Mn2+ + 8H2O+5Cl2
This is the balanced equation in an acidic medium.
That is 8, right side.
To know more about redox reaction follow the link:
https://brainly.in/question/9854479
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<h3>
Answer:</h3>
0.111 J/g°C
<h3>
Explanation:</h3>
We are given;
- Mass of the unknown metal sample as 58.932 g
- Initial temperature of the metal sample as 101°C
- Final temperature of metal is 23.68 °C
- Volume of pure water = 45.2 mL
But, density of pure water = 1 g/mL
- Therefore; mass of pure water is 45.2 g
- Initial temperature of water = 21°C
- Final temperature of water is 23.68 °C
- Specific heat capacity of water = 4.184 J/g°C
We are required to determine the specific heat of the metal;
<h3>Step 1: Calculate the amount of heat gained by pure water</h3>
Q = m × c × ΔT
For water, ΔT = 23.68 °C - 21° C
= 2.68 °C
Thus;
Q = 45.2 g × 4.184 J/g°C × 2.68°C
= 506.833 Joules
<h3>Step 2: Heat released by the unknown metal sample</h3>
We know that, Q = m × c × ΔT
For the unknown metal, ΔT = 101° C - 23.68 °C
= 77.32°C
Assuming the specific heat capacity of the unknown metal is c
Then;
Q = 58.932 g × c × 77.32°C
= 4556.62c Joules
<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>
- We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.
4556.62c Joules = 506.833 Joules
c = 506.833 ÷4556.62
= 0.111 J/g°C
Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C
