What is the ph of a buffer that is 0. 08 m naa and 0. 1 ha. The ka of ha = 1. 0 x 10^-4
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Answer:
4.03dm³
Explanation:
The reaction expression is given as:
3H₂ + N₂ → 2NH₃
Volume of hydrogen = 12dm³
AT rtp:
1 mole of gas occupies volume of 22.4dm³
x mole of hydrogen will occupy a volume of 12dm³
Number of moles of hydrogen = = 0.54mole
From the balanced reaction equation:
3 mole of hydrogen gas combines with 1 mole of Nitrogen gas
0.54 mole of hydrogen as will therefore combine with = 0.18moles of nitrogen gas
Since ;
1 mole of gas occupies a volume of 22.4dm³
0.18moles of Nitrogen gas will occupy 0.18 x 22.4 = 4.03dm³