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algol13
3 years ago
10

How many moles of MgS2O3 are in 205 g of the compound?

Chemistry
2 answers:
Agata [3.3K]3 years ago
6 0
Mol =   \frac{mass}{molar mass}

Mass of the compound = 205 g
Molar Mass of compound = ((24) + (2 * 32) + (3 * 16))
                                         =  136 g / mol

∴ # mols in 205g =  \frac{205 g}{136 g / mol}
                           = 1.507 mol 

serg [7]3 years ago
6 0
<span>Since a mol of MgS2O3 weights 136gr (the same weight as the atomic weight for that compound, but measured in gr), you just have to divide the mass you have (205) by that one (136), resulting in 1'514 moles.</span>
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If a compound contains an anion which is the conjugate base of a weak acid, the addition of hydronium ion solubility does what ?
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Answer:

The addition of hydronium ion increases the solubility of the salt

Explanation:

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MA ----> M+ + A-

The anion which is a conjugate base will be removed from the solution with the addition of a hydronium ion from an acid: 2H3O+ + 2A- ----> 2HA + 2H20

This distorts the reaction equilibrium. According to Le Chatelier’s principle, more MA will dissolve until in order to restore the previous equilibrium of the reaction.

Therefore, an acidic pH increases the solubility of almost all sparingly soluble salts whose anion is the conjugate base of a weak acid.

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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
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<u>Answer:</u> The amount of energy released per gram of B_5H_9 is -71.92 kJ

<u>Explanation:</u>

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

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3 years ago
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