Given :
2.5 mole of Sulfuric acid .
To Find :
Mass of sodium hydroxide will completely neutralize 2.5 mol of sulfuric acid
Solution :
Let us assume volume of water be 1 L .
Now , we know , to neutralize 1 mole of sulfuric acid we need 2 moles of NaOH .
So , for 2.5 mole sulfuric acid required 5 mole of NaOH .
Moles of NaOH ,
Molecular mass of NaOH , M.M = 58.44 g/mol .
Mass of 5 moles of NaOH :
Hence , this is the required solution .
Answer:
12.8 g of must be withdrawn from tank
Explanation:
Let's assume gas inside tank behaves ideally.
According to ideal gas equation-
where P is pressure of , V is volume of
, n is number of moles of
, R is gas constant and T is temperature in kelvin scale.
We can also write,
Here V, T and R are constants.
So, ratio will also be constant before and after removal of
from tank
Hence,
Here, and
So,
So, moles of must be withdrawn = (0.66 - 0.26) mol = 0.40 mol
Molar mass of = 32 g/mol
So, mass of must be withdrawn =
For a voltaic cell consisting of chromium, an electrode dipped in a 1.20 M chromium (III) nitrate solution and a tin electrode dipped in a 0.400 M tin (II) nitrate solution, the cell potential at 298 K is mathematically given as
Ecell = 0.577 V
<h3 /><h3>What is the cell potential at 298 K?</h3>Generally, the equation for the Oxidation and Reduction is mathematically given as
Cr(s) ------------------ Cr+3(aq) + 3e- ] x 2 ...O
Sn+2(aq) + 2e- ------------ Sn(s) ] x 3 ...R
Reaction
2 Cr(s) + 3 Sn+2(aq) --------------- 2 Cr+3(aq) + 3 Sn(s)
Therefore
Eicell = - 0.14 - ( - 0.74)
Eicell = 0.60
In conclusion
Ecell = 0.577 V
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