Answer:
D. Ultraviolet light should have a short wavelength, not a long wavelength.
Explanation:
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Answer:
104.84 moles
Explanation:
Given data:
Moles of Boron produced = ?
Mass of B₂O₃ = 3650 g
Solution:
Chemical equation:
6K + B₂O₃ → 3K₂O + 2B
Number of moles of B₂O₃:
Number of moles = mass/ molar mass
Number of moles = 3650 g/ 69.63 g/mol
Number of moles = 52.42 mol
Now we will compare the moles of B₂O₃ with B from balance chemical equation:
B₂O₃ : B
1 : 2
52.42 : 2×52.42 = 104.84
Thus from 3650 g of B₂O₃ 104.84 moles of boron will produced.
Answer
- continuous removal of PH3
- adding more of P into the system
Explanation:
In the reaction P4(g)+6H2(g) ⇌ 4PH3(g);
- The effect of temperature on equilibrium has to do with the heat of reaction. Recall that for an endothermic reaction, heat is absorbed in the reaction, and the value of ΔH is positive. Thus, for an endothermic reaction, we can picture heat as being a reactant:
heat+A⇌BΔH=+
- Since the reaction is endothermic reaction, heat is a absorbed. Decreasing the temperature will shift the equilibrium to the left, while increasing the temperature will shift the equilibrium to the right forming more of PH3.
- According to Le Chatelier’s principle, adding additional reactant to a system will shift the equilibrium to the right, towards the side of the products. In the same Way, reducing the concentration of the product will also shift equilibrium to the right continually forming PH3 as it is removed.
M= #moles / L
4.35/.75 = 5.6
Explanation:
Formula for work done is as follows.
W = 
where, k = proportionality constant = 
d = separation distance = 0.45 nm = 
Now, we will put the given values into the above formula and calculate work done as follows.
W =
= ![\frac{-[8.99 \times 10^{9} Jm/C^{2} \times 3.2 \times 10^{-19} C \times -3.2 \times 10^{-19} C]}{0.25 \times 10^{-9} m}](https://tex.z-dn.net/?f=%5Cfrac%7B-%5B8.99%20%5Ctimes%2010%5E%7B9%7D%20Jm%2FC%5E%7B2%7D%20%5Ctimes%203.2%20%5Ctimes%2010%5E%7B-19%7D%20C%20%5Ctimes%20-3.2%20%5Ctimes%2010%5E%7B-19%7D%20C%5D%7D%7B0.25%20%5Ctimes%2010%5E%7B-9%7D%20m%7D)
= 
Thus, we can conclude that work required to increase the separation of the two ions to an infinite distance is 
.
