Which of the following is an oxidation-reduction reaction? Which of the following is an oxidation-reduction reaction? Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) HCl(aq) + LiOH(aq) → LiCl(aq) + H2O(l) NaCl(aq) + AgNO3(aq) → AgCl(s) + NaNO3(aq) Pb(C2H3O2)2(aq) + 2 NaCl(aq) → PbCl2(s) + 2 NaC2H3O2(aq) All of the above are oxidation-reduction reactions.
Answers
Answer:
NONE OF THE ABOVE
Explanation:
None of the above are examples of an oxidation - reduction or a redox reaction . This is because there is no change in the oxidation state of any of the elements in the reaction when the reaction happens .
- For a redox reaction , transfer of electrons must take place.
- The oxidation states must increase or decrease in atleast 2 of the elements of a compound.
- For example :
⇒
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Answer:
139.33 g of magnesium chloride, MgCl2.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(aq)
Next, we shall determine the mass of Mg that reacted and the mass of MgCl2 from the balanced equation.
This is illustrated below:
Molar mass of Mg = 24 g/mol
Mass of Mg from the balanced equation = 1 x 24 = 24 g
Molar mass of MgCl2 = 24 + (2x35.5) = 95 g/mol
Mass of MgCl2 from the balanced equation = 1 x 95 = 95 g
From the balanced equation above,
24 g of Mg reacted to produce 95 g of MgCl2.
Finally, we shall determine the mass of MgCl2 produced by reacting 35.2 g of Mg.
This can be obtained as follow:
From the balanced equation above,
24 g of Mg reacted to produce 95 g of MgCl2.
Therefore, 35.2 g of Mg will react to produce = (35.2 x 95)/24 = 139.33 g of MgCl2.
From the calculations made above, 139.33 g of magnesium chloride, MgCl2 were produced.
Answers
Answers
Answer:
The second experiment (reversible path) does more work
Explanation:
Step 1:
A piston confines 0.200 mol Ne(g) in 1.20L at 25 degree °C
(a) The gas is allowed to expand through an additional 1.20 L against a constant of 1.00atm
Irreversible path: w =-Pex*ΔV
⇒ with Pex = 1.00 atm
⇒ with ΔV = 1.20 L
W = -(1.00 atm) * 1.20 L
W = -1.20L*atm *101.325 J /1 L*atm = -121.59 J
(b) The gas is allowed to expand reversibly and isothermally to the same final volume.
W = -nRTln(Vfinal/Vinitial)
⇒ with n = the number of moles = 0.200
⇒ with R = gas constant = 8.3145 J/K*mol
⇒ with T = 298 Kelvin
⇒ with Vfinal/Vinitial = 2.40/1.20 = 2
W = -(0.200mol) * 8.3145 J/K*mol *298K *ln(2.4/1.2)
W = -343.5 J
The second experiment (reversible path) does more work