An oxidation-reduction (redox<span>) </span>reaction<span> is a type of chemical </span>reaction<span> that involves a transfer of electrons between two species. An oxidation-reduction </span>reaction<span> is any chemical </span>reaction<span> in which the oxidation number of a molecule, atom, or ion changes by gaining or losing an electron.</span>
To find molarity
1) number of mol of solute.
Solute is HCl.
M(HCl)= 1.0+35.5 =36.5 g/mol
25g *1 mol/36.5 g = 25/36.5 mol HCl
2) Molarity is number of mole of the solute in 1 L solution.
150 mL = 0.150 L
(25/36.5 mol HCl )/(0.150 L) = 25/(36.5*0.150) ≈ 4.57≈4.6 mol/L
In a particular experiment, the per cent yield is 79.0%. This means that in this experiment, a 7.90-g sample of fluorine yields is 7g of SF6.
<h3>How is Sulphur hexafluoride formed?</h3>
Sulfur Hexafluoride is a disparity agent formed of an inorganic fluorinated inert gas comprised of six fluoride atoms bound to one sulfur atom, with possible diagnostic activity upon imaging.
Thus, a sample of fluorine yields 7g of SF6.
To learn more about Sulfur Hexafluoride click here;
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Answer:
1) The value of Kc:
C. remains the same.
2) The value of Qc:
A. is greater than Kc.
3) The reaction must:
B. run in the reverse direction to restablish equilibrium.
4) The concentration of N2 will:
B. decrease.
Explanation:
Hello,
In this case, by means of the Le Chatelier's principle which is based on the shift a chemical reaction could have under some modifications, we have:
1) The value of Kc:
C. remains the same, since it just depend the reaction's thermodynamics as it is computed via:
2) The value of Qc:
A. is greater than Kc, since the reaction quotient is:
![Qc=\frac{[N_2][H_2]^3}{[NH_3]^2}](https://tex.z-dn.net/?f=Qc%3D%5Cfrac%7B%5BN_2%5D%5BH_2%5D%5E3%7D%7B%5BNH_3%5D%5E2%7D)
Thus, the lower the concentration of ammonia, the higher Qc, making Qc>Kc.
3) The reaction must:
B. run in the reverse direction to restablish equilibrium, since ammonia was withdrawn and should be regenerated to reach the equilibrium.
4) The concentration of N2 will:
B. decrease, since less reactant is forming the products.
Best regards.
We need (i) the stoichiometric equation, and (ii) the equivalent mass of dihydrogen.
Explanation:
1
2
N
2
(
g
)
+
3
2
H
2
(
g
)
→
N
H
3
(
g
)
11.27
g
of ammonia represents
11.27
⋅
g
17.03
⋅
g
⋅
m
o
l
−
1
=
?
?
m
o
l
.
Whatever this molar quantity is, it is clear from the stoichiometry of the reaction that 3/2 equiv of dihydrogen gas were required. How much dinitrogen gas was required?
