Because subatomic particles ARE what make up atoms.
Answer:
[IBr] = 0.049 M.
Explanation:
Hello there!
In this case, according to the balanced chemical reaction:
It is possible to set up the following equilibrium expression:
![K=\frac{[IBr]^2}{[I_2][Br_2]} =0.0110](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BIBr%5D%5E2%7D%7B%5BI_2%5D%5BBr_2%5D%7D%20%3D0.0110)
Whereas the the initial concentrations of both iodine and bromine are 0.50 M; and in terms of 
(reaction extent) would be:
Which can be solved for to obtain two possible results:
Whereas the correct result is 0.0245 M since negative results does not make any sense. Thus, the concentration of the product turns out:
![[IBr]=2x=2*0.0249M=0.049M](https://tex.z-dn.net/?f=%5BIBr%5D%3D2x%3D2%2A0.0249M%3D0.049M)
Regards!
Answer:
The correct answer is option c. transfer of electrons from Mg to O.
Explanation:
Hello!
Let's solve this!
When Magnesium (Mg) reacts with oxygen (O), magnesium oxide is formed.
This reaction is spontaneous and occurs with oxidation number +2 of magnesium and oxidation number -2 of oxygen. It is an ionic union, so magnesium transfers its electrons to oxygen.
We conclude that the correct answer is option c. transfer of electrons from Mg to O.
1. you need a periodic table and find the atomic mass of Cu (copper), S (sulfur) and O (oxygen). The atomic mass is the number in the box that corresponds to the element and have several decimal places.
2. atomic mass of
Cu = 63.546
S = 32.065
O = 15.9994
3. Then according to the formula of the compound, you add as many time the atomic mass of each element as subindex in the formula and add all the values together to calculate the molecular mass of the compound in grams.
4. 63.546 g + 32.065 g + ( 4 x <span>15.9994) = 159.609 g
5. this value </span><span>159.609 g is the mass in grams of one mol of CuSO4
6 the problem is asking not for the mass of one mole but the mass of 3.65 moles of CuSO4
7 then you have the multiply the value of one mol by the number of moles that the problem is asking you
8. </span><span>159.609 g x 3.65 = 582.571 g
</span>
9 the answer to the problem will be
"there are 582.571 g of CuSO4 in 3.65 moles of CuSO4"
