Answer:
0.1313 g.
Explanation:
- It is known that at STP, 1.0 mole of ideal gas occupies 22.4 L.
- Suppose that hydrogen behaves ideally and at STP conditions.
<u><em>Using cross multiplication:</em></u>
1.0 mol of hydrogen occupies → 22.4 L.
??? mol of hydrogen occupies → 1.47 L.
∴ The no. of moles of hydrogen that occupies 1.47 L = (1.0 mol)(1.47 L)/(22.4 L) = 6.563 x 10⁻² mol.
- Now, we can get the no. of grams of hydrogen in 6.563 x 10⁻² mol:
<em>The no. of grams of hydrogen = no. of hydrogen moles x molar mass of hydrogen</em> = (6.563 x 10⁻² mol)(2.0 g/mol) = <em>0.1313 g.</em>
Answer:
c.hg cannot be cracked for fractional distillation as there is only one of each
Explanation:
The enthalpy change : -196.2 kJ/mol
<h3>Further explanation </h3>
The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation
The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)
(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)
The value of ° H ° can be calculated from the change in enthalpy of standard formation:
∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)
Reaction
2 H₂O₂(l)-→ 2 H₂O(l) + O₂(g)
∆H ° rxn = 2. ∆Hf ° H₂O - 2. ∆Hf °H₂O₂
![\tt \Delta H_{rxn}=2.(-285.8)-2.(-187.8)\\\\\Delta H_{rxn}=-571.6+375.4=-196.2~kJ/mol\rightarrow \Delta Hf~O_2=0](https://tex.z-dn.net/?f=%5Ctt%20%5CDelta%20H_%7Brxn%7D%3D2.%28-285.8%29-2.%28-187.8%29%5C%5C%5C%5C%5CDelta%20H_%7Brxn%7D%3D-571.6%2B375.4%3D-196.2~kJ%2Fmol%5Crightarrow%20%5CDelta%20Hf~O_2%3D0)
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