Answer:
Mg + Fe(NO₃)₂ —> Fe + Mg(NO₃)₂
Explanation:
The activity series helps us to easily define whether or not a reaction will occur.
Elements at the top of the activity series are highly reactive and will always displace those at the bottom of the series in any reaction.
With the above information in mind, let us answer the questions given above.
Ag + NaNO₃ —> Na + AgNO₃
The above reaction will not occur because Na is higher than Ag in the activity series. Thus, Ag cannot displace Na from solution.
Pb + Mg(NO₃)₂ —> Pb(NO₃)₂ + Mg
The above reaction will not occur because Mg is higher than Pb in the activity series. Thus, Pb cannot displace Mg from solution.
Mg + Fe(NO₃)₂ —> Fe + Mg(NO₃)₂
The above reaction will occur because Mg is higher than Fe in the activity series. Thus, Mg will displace Fe from solution.
Cu + Mg(NO₃)₂ —> Cu(NO₃)₂ + Mg
The above reaction will not occur because Mg is higher than Cu in the activity series. Thus, Cu cannot displace Mg from solution.
From the above illustration, only
Mg + Fe(NO₃)₂ —> Fe + Mg(NO₃)₂
Will occur.
The moles of the gas in the sample is 0.391 moles
calculation
by use of ideal gas equation, that is Pv=nRT
where n is number of moles
P(pressure)= 660 mmhg
R(gas constant) = 62.364 l.mmhg/mol.K
T(temperature)= 25 +273 = 298 k
by making n the subject of the formula
n= Pv/ RT
n is therefore= (660mm hg x11 L)/( 62.364 L.mmhg/mol.k x298 K) = 0.391 moles
Answer:
0.26 mol
Explanation:
using general gas equation
PV=nRT
V=4.1litre= 4.1 dm³
P= 1.78 atm
R= 0.0821
PUT VALUES
It is a good thing that you already have answered the first question. Now, moving on to the second question, there exist an equation for the neutralization of acid by a base that is shown below,
M₁V₁ = M₂V₂
Now, all the variables in the equation are given except for our unknown which is the V₂. Substituting the known values from the given above,
(0.1 M)(25 mL) = (0.05 M)(V₂)
The value of V₂ from the equation above is 50 mL. Therefore, 50 mL of 0.05 M NaOH solution will be needed to completely react with HNO3.
1) 0.1 M NaCl
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