Mass spectrum of Dodecane will give following information.
1 ) Molecular Peak or Parent Peak:
The Parent peak will appear at m/z = 170. The intensity of this peak will be very weak.
2) Fragments:
Usually the fragments of such long chain alkanes appear with spacing of 14 amu, Hence, the peaks in dodecane will be as follow,
170 - 156 - 142 - 128 - 114 - 100 - 86 - 72 - 58 - 44 - 30 - 16
3) Base Peak:
Most probably the Base peak will appear at m/z = 57. This peak is due to the formation of tertiary butyl cation as the intensity mainly depends upon the stability of cation. So this cation might form due to rearrangment giving the intensity of 100%.
Answer:
290.82g
Explanation:
The equation for the reaction is given below:
2Al + 3H2SO4 -> Al2(SO4)3 + 3H2 now, let us obtain the masses of H2SO4 and Al2(SO4)3 from the balanced equation. This is illustrated below:
Molar Mass of H2SO4 = (2x1) + 32 + (16x4) = 2 + 32 +64 = 98g/mol
Mass of H2SO4 from the balanced equation = 3 x 98 = 294g
Molar Mass of Al2(SO4)3 = (2x27) + 3[32 + (16x4)]
= 54 + 3[32 + 64]
= 54 + 3[96] = 54 + 288 = 342g
Now, we can obtain the mass of aluminium sulphate formed by doing the following:
From the equation above:
294g of H2SO4 produced 342g of Al2(SO4)3.
Therefore, 250g of H2SO4 will produce = (250 x 342)/294 = 290.82g of Al(SO4)3
Therefore, 290.82g of aluminium sulphate (Al(SO4)3) is formed.
Answer:
The enthalpy of atomization is the enthalpy change that accompanies the total separation of all atoms in a chemical substance. This is often represented by the symbol ΔₐₜH or ΔHₐₜ. All bonds in the compound are broken in atomization and none are formed, so enthalpies of atomization are always positive.
Explanation:
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The balanced chemical equation is written as:
<span>CsF(s) + XeF6(s) ------> CsXeF7(s)
We are given the amount of </span>cesium fluoride and <span>xenon hexafluoride used for the reaction. We need to determine first the limiting reactant to proceed with the calculation. From the equation and the amounts, we can say that the limiting reactant would be cesium fluoride. We calculate as follows:
11.0 mol CsF ( 1 mol </span>CsXeF7 / 1 mol CsF ) = 11.0 mol <span>CsXeF7</span>
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Hope this helps you!
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