NaH(s)+ H2O (l)=>NaOH(aq)+H2(g)
You want to calculate the mass of NaH, I assume. Otherwise, the question isn't clear. It simply says calculate the mass(??)
So, calculate the moles of H2 gas that satisfy the conditions of 982 ml at 28ºC and 765 torr. But you must subtract the vapor pressure of water at 28º to get the actual pressure of the H2 gas. So, the actual conditions are 982 ml (0.982 L) and 301 K and 765-28 = 737 torr.
PV = nRT
n = PV/RT = (737 torr)(0.982 L)/(62.4 L-torr/Kmol)(301 K)
n = 0.0385 moles H2
moles NaH needed = 0.0385 moles H2 x 1 mole NaH/mole H2 = 0.0385 moles NaH required
mass of NaH needed = 0.0385 moles x 24 g/mole = 0.925 g NaH
Brainliest Please :)
The answer is double displacement is the answer
Balanced Half reactions are:
At anode 2
==> Cl₂+ 
+ H₂O ==> 
+ 2
+
At Cathode: 2 + ==> H₂
Since the question states that you are using an aqueous solution of MnCl₂, so ions will have present are, H₂O, , 
and
Now at Anode reaction will occur as given:
2 ==> Cl₂+ + H₂O ==> + 2 + (will occur)
At Cathode:
2 + ==> H₂ (will occur)
At Cathode:
+ ==> Mn (This reaction will not occur)
The deposition of solid Mn will not occur because in aqueous solution, will be reduced before .
The reduction potentials for is zero whereas reduction potential for is - 1.18V.
The reduction potential of a species is its tendency to gain electrons and get reduced. It is measured in millivolts or volts. Larger positive values of reduction potential are indicative of a greater tendency to get reduced.
To learn more about the half reaction please click on the link brainly.com/question/13186640
#SPJ4
