Number of moles = number of atoms/Avogadro's number
= (4.8x10^25)/(6.02x10^23)
=79.7 mol (3sf)
Answer:
"HEMOGLOBINA. Las características de la hemoglobina (Hb) como amortiguador están íntimamente vinculadas a la capacidad de disociación del grupo imidazólico del aminoácido histidina unido al hierro que contiene el grupo hemo (ver figura inferior)."
Link: http://www.ehu.eus/biomoleculas/buffers/buffer4.htm#:~:text=HEMOGLOBINA&text=Las%20caracter%C3%ADsticas%20de%20la%20hemoglobina,hemo%20(ver%20figura%20inferior).
A) NaCl = 58.44 g/mol
number of moles : 2.14 / 58.44 = 0.0366 moles
V = n / M => 0.0366 / 0.270 = 0.1355 L
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b) C2H6<span>O = 46.06 g/mol
</span>number of moles : 4.30 / 46.06 = 0.0933 moles
<span>V = n / M => 0.0933 / 1.50 = 0.0622 L
</span>___________________________________________
c) CH3COOH = 60.0 g/mol
number of moles : 0.85 / 60.0= 0.0141 moles
V = n / M => <span> 0.0141 </span>/ 0.30 = 0.047 L
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hope this helps!
Answer:
you know that they will be a displacement reaction that will form a barium salt:
Ba(NO3)2+ 2NaCl--> BaCl2 + 2NaNO3
So now that we have that formula and the molecular weight we can determine how much salt will be made. So here we convert the grams to moles
(42.3g Ba(NO3)2)*(1 mole/261.34g) = 0.16185 mol
In the molecular formula we know that 1 mole of Barium nitrate will create 1 mole of Barium chloride, so in this case (in a perfect world) you should get 0.16185 mole of barium chloride (208.23 g/mol) that we then have to convert to grams.
(0.16185 mol BaCl2) * ( 208.23 g/mol) = 33.7037 g of Barium Chloride (rounded to 3 significant digits = 33.7g)
The question is incomplete, here is the complete question:
At a certain temperature this reaction follows second-order kinetics with a rate constant of 14.1 M⁻¹s⁻¹
Suppose a vessel contains SO₃ at a concentration of 1.44 M. Calculate the concentration of SO₃ in the vessel 0.240 seconds later. You may assume no other reaction is important. Round your answer to 2 significant digits.
<u>Answer:</u> The concentration of 
in the vessel after 0.240 seconds is 0.24 M
<u>Explanation:</u>
For the given chemical equation:
The integrated rate law equation for second order reaction follows:
![k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B1%7D%7Bt%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%5BA%5D_o%7D%5Cright%29)
where,
k = rate constant = 
t = time taken= 0.240 second
[A] = concentration of substance after time 't' = ?
![[A]_o](https://tex.z-dn.net/?f=%5BA%5D_o)
= Initial concentration = 1.44 M
Putting values in above equation, we get:
![14.1=\frac{1}{0.240}\left (\frac{1}{[A]}-\frac{1}{1.44}\right)](https://tex.z-dn.net/?f=14.1%3D%5Cfrac%7B1%7D%7B0.240%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B1.44%7D%5Cright%29)
![[A]=0.245M](https://tex.z-dn.net/?f=%5BA%5D%3D0.245M)
Hence, the concentration of in the vessel after 0.240 seconds is 0.24 M
