Answer:
3 moles of CO are needed
Explanation:
Given data:
Number of moles of CO used = ?
Mass of Fe produced = 112 g
Solution:
Chemical equation:
Fe₂O₃ + 3CO → 2Fe + 3CO₂
Number of moles of Fe:
Number of moles = mass/ molar mass
Number of moles = 112 g/ 55.85 g/mol
Number of moles = 2.00 mol
Now we will compare the moles of iron and CO.
Fe : CO
2 : 3
Thus, 3 moles of CO are needed.
How do the number of electrons in the second energy shell of an atom change going across period 2 in the periodic table?
The answer is: A. The number of electrons increases by 1, from left to right.
B is wrong because atoms in period 2 DO have a second energy shell. For C, the number of electrons in the second energy shell is NOT full. For D, it is wrong to state that the number of electrons increases by 1, from right to left.
Answer:
Explanation:
An element is a _distinct____ substance represented by a __chemical____ symbol. Chemical symbols have _one__ or _two__ letters. The first letter of a chemical symbol is always a _Capital_____ letter and the second letter is always a _small_____ letter.
Elements are distint substances that cannot be split up into simpler substances. Such substances are made up of one kind of atom. Each of them is usually symbolised by a capital letter or a capital letter followed by a small letter derieved from the English or latin or greek name of the element.
Answer:
C47N9O18 is the empirical formula of this compound.
Explanation:
Answer:
b. 
Explanation:
Hello there!
In this case, given the ionization reaction of HClO as weak acid:
We can write the equilibrium expression as shown below:
![Ka=3.5x10^{-8}=\frac{[H^+][ClO^-]}{[HClO]}](https://tex.z-dn.net/?f=Ka%3D3.5x10%5E%7B-8%7D%3D%5Cfrac%7B%5BH%5E%2B%5D%5BClO%5E-%5D%7D%7B%5BHClO%5D%7D)
In such a way, via the definition of x as the reaction extent, we can write:
![3.5x10^{-8}=\frac{x^2}{[HClO]}](https://tex.z-dn.net/?f=3.5x10%5E%7B-8%7D%3D%5Cfrac%7Bx%5E2%7D%7B%5BHClO%5D%7D)
As long as Ka<<<<1 so that the x on the bottom can be neglected. Thus, we solve for x as shown below:
And finally the percent dissociation:
![\% diss=\frac{x}{[HClO]} *100\%\\\\\% diss=\frac{7.94x10^{-5}M}{0.18}*100\% \\\\\% diss =0.044\%=4.4x10^{-2}\%](https://tex.z-dn.net/?f=%5C%25%20diss%3D%5Cfrac%7Bx%7D%7B%5BHClO%5D%7D%20%2A100%5C%25%5C%5C%5C%5C%5C%25%20diss%3D%5Cfrac%7B7.94x10%5E%7B-5%7DM%7D%7B0.18%7D%2A100%5C%25%20%5C%5C%5C%5C%5C%25%20diss%20%3D0.044%5C%25%3D4.4x10%5E%7B-2%7D%5C%25)
Which is choice b.
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