Volume of H2 produced = 57.6576 L
<h3>Further explanation</h3>
Given
23.17 g Be
Required
Volume of H2
Solution
Reaction
Be(s)+H2O(g)→BeO(s)+H2(g)
mol Be :
= 23.17 g : 9 g/mol
= 2.574
From the equation, mol H2 : mol Be = 1 : 1, so mol H2 = 2.574
Volume H2(assumed at STP, 1 mol=22.4 L) :
= 2.574 x 22.4 L
= 57.6576 L
Answer:1-methoxy-2,4-dinitrobenzene
Explanation:
The nitro groups are strongly electron withdrawing and promote nucleophilic substitution reactions where one of the original substituents is removed and replaced by a strong nucleophile such as the methoxy group. The mechanism of the reaction is attached below. The electron withdrawing nitrogroup assists the formation of the intermediate in the reaction as shown.
D Earth's core contains mostly iron and nickel
<u>Answer:</u> The
for the reaction is 54.6 kJ/mol
<u>Explanation:</u>
For the given balanced chemical equation:

We are given:

- To calculate
for the reaction, we use the equation:
![\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20G_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20G_f%28reactant%29%5D)
For the given equation:
![\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28COCl_2%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28CO_2%29%7D%29%2B%281%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28CCl_4%29%7D%29%5D)
Putting values in above equation, we get:
![\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%28-204.9%29%29-%28%281%5Ctimes%20%28-394.4%29%29%2B%281%5Ctimes%20%28-62.3%29%29%29%5D%5C%5C%5CDelta%20G%5Eo_%7Brxn%7D%3D46.9kJ%3D46900J)
Conversion factor used = 1 kJ = 1000 J
- The expression of
for the given reaction:

We are given:

Putting values in above equation, we get:

- To calculate the Gibbs free energy of the reaction, we use the equation:

where,
= Gibbs' free energy of the reaction = ?
= Standard gibbs' free energy change of the reaction = 46900 J
R = Gas constant = 
T = Temperature = ![25^oC=[25+273]K=298K](https://tex.z-dn.net/?f=25%5EoC%3D%5B25%2B273%5DK%3D298K)
= equilibrium constant in terms of partial pressure = 22.92
Putting values in above equation, we get:

Hence, the
for the reaction is 54.6 kJ/mol