Answer:
5746.0 mL.
Explanation:
We can use the general law of ideal gas:<em> PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
If n and P are constant, and have two different values of V and T:
<em>V₁T₂ = V₂T₁</em>
<em></em>
V₁ = 6193.0 mL, T₁ = 62.3°C + 273 = 335.3 K.
V₂ = ??? mL, T₂ = 38.1°C + 273 = 311.1 K.
<em>∴ V₂ = V₁T₂/T₁ </em>= (6193.0 mL)(311.1 K)/(335.3 K) = <em>5746.0 mL.</em>
If you would’ve attached a picture I’m sure it would’ve been a lot easier.
Answer:
Cu(NO3)2(aq)+Pb(s) ⇌ Pb(NO3)2(aq)+Cu(s)
Explanation:
If we look at the both reactions closely, we will quickly discover that the reaction CuSO4(aq)+Pb(s) ⇌ PbSO4(s)+Cu(s) involves PbSO4.
The compound PbSO4 is insoluble in water and sinks to the bottom of the reaction vessel. When this occurs, the concentration of Pb^2+ becomes low. This will bring about a low voltage in the cell.
On the other hand, Pb(NO3)2 is soluble in water hence the cell voltage in this case is higher than the former.
<em><u />the correct option is <u>c ) 10,000...</u>
an acid which is more acidic than 6 pH will be of pH 5 acid because if one acid is stronger than the other acid than its magnitude of strength is 1000... it is constant always...</em>
Answer:
2) Add a solution of NaBr
Explanation:
Lead (II) bromide is an inorganic powdery substance that has a solubility in water of 0.973 g/100 mL at 20°C. It is insoluble in alcohol but is soluble in alkali, ammonia, NaBr, and KBr
PbBr₂ is slightly soluble in ammonia, and it reacts with NaOH to produce Pb(OH)₂ and NaBr
Therefore, the best solution for dissolving PbBr₂(s) is NaBr
