The answer for the first one is the second option.
The answer for the second one is the last option.
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The answer is A
explanation
Explanation:
(a) The given data is as follows.
T = = (55 + 273) K
= 328 K
P = 35 bar = 3500 kPa
Let us assume that moles of ethylene present are 1 kmol and according to the ideal gas equation, PV = nRT.
Or, V =
=
= 0.779
Hence, the volume occupied by 18 kg of ethylene at and 35 bar is 0.779 .
(b) The given data is as follows.
V = 0.25
T = = (50 + 273) K
= 323 K
P = 115 bar = 11500 kPa
Using ideal gas equation first, we will calculate its moles as follows.
n =
=
= 1.07 mol
Since, moles =
Hence, mass of ethylene will be calculated as follows.
moles =
1.07 mol =
mass = 29.96 g
Therefore, mass of given ethylene is 29.96 g.
HCO3- + H2O ---><--- H2CO3 + OH-
<span>HCO3- + H2O ----><--- CO3-2 + H3O+ </span>
<span>Dihydrogen Phosphate is H2PO4-1 </span>
<span>H2PO4-1 + H2O ---><--- H3PO4 + OH- </span>
<span>H2PO4-1 + H2O ---><--- HPO4-2 + H3O+ </span>
<span>HIDROGEN PHOSPHATE IS HPO4-2 and for it you've got: </span>
<span>HPO4-2 + H2O ---><--- H2PO4-1 + OH- </span>
<span>HPO4-2 + H2O ---><--- PO4-3 + H3O</span>