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Pentaborane-9, B5H9, is a colorless, highly reactive liquid that will burst into flame when exposed to oxygen. The reaction is 2

mina [271]

Answer: The amount of energy released per gram of $B_5H_9$ is -71.92 kJ

Explanation:

For the given chemical reaction:

$2B_5H_9(l)+12O_2(g)\rightarrow 5B_2O_3(s)+9H_2O(l)$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(5\times \Delta H^o_f_{(B_2O_3(s))})+(9\times \Delta H^o_f_{(H_2O(l))})]-[(2\times \Delta H^o_f_{(B_5H_9(l))})+(12\times \Delta H^o_f_{(O_2(g))})]$

Taking the standard enthalpy of formation:

$\Delta H^o_f_{(B_2O_3(s))}=-1271.94kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.83kJ/mol\\\Delta H^o_f_{(B_5H_9(l))}=73.2kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(5\times (1271.94))+(9\times (-285.83))]-[(2\times (73.2))+(12\times (0))]\\\\\Delta H^o_{rxn}=-9078.57kJ$

We know that:

Molar mass of pentaborane -9 = 63.12 g/mol

By Stoichiometry of the reaction:

If 2 moles of $B_5H_9$ produces -9078.57 kJ of energy.

Or,

If $(2\times 63.12)g$ of $B_5H_9$ produces -9078.57 kJ of energy

Then, 1 gram of $B_5H_9$ will produce = $\frac{-9078.57kJ}{(2\times 63.12)}\times 1g=-71.92kJ$ of energy.

Hence, the amount of energy released per gram of $B_5H_9$ is -71.92 kJ

8 0

3 years ago

3. Heating the copper product at too high a temperature in an oxygen atmosphere results in the formation of copper( )oxde, . Wri

olya-2409 [2.1K]

First, write the equation:
Cu + O2 -> CuO
Now, balance:
2 Cu + O2 -> 2 CuO

6 0

3 years ago

Atoms with an electric charge come charged by gaining or losing

dexar [7]

Is the atoms of electric

6 0

3 years ago

A government laboratory wants to determine whether water in a certain city has any traces of fluoride and whether the concentrat

nexus9112 [7]

Answer: The given sample of water is not safe for drinking.

Explanation:

We are given:

Concentration of fluorine in water recommended = 4.00 ppm

ppm is the amount of solute (in milligrams) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of fluorine in water, we use the equation:

$\text{ppm}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^6$

Both the masses are in grams.

We are given:

Mass of fluorine = $0.152mg=0.152\times 10^{-3}g$ (Conversion factor: 1 g = 1000 mg)

Mass of water = 5.00 g

Putting values in above equation, we get:

$\text{ppm of fluorine in water}=\frac{0.152\times 10^{-3}}{5}\times 10^6\\\\\text{ppm of fluorine in water}=30.4$

As, the calculated concentration is greater than the recommended concentration. So, the given sample of water is not safe for drinking.

Hence, the given sample of water is not safe for drinking.

6 0

3 years ago

The given potentials are observed for a calomel electrode: E ° = 0.268 V and E ( saturated KCl ) = 0.241 V . Use this informatio

Anvisha [2.4K]

Answer: The E for Silver-silver Chloride electrode = 0.287 V

Explanation:

Silver/Silver Chloride (Ag/AgCl) with a value for E° that is actually +0.222 V or approximately 0.23 V has the actual potential of the half-cell prepared in this way as +0.197 V vs SHE, (Standard Hydrogen Electrode) which arises because in addition to KCl, there is the contribuion of AgCl to the chloride activity, which isn't exactly unity.

Therefore, the E for the Ag/AgCl electrode would approximately equal 0.287 V

4 0

3 years ago