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3 years ago

14

172 °C = __________ K (Round off to the nearest whole number)

1 answer:

NNADVOKAT [17]3 years ago

8 0

-101, K=C+273 so in this you would subtract 273 from 172

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4 years ago

Read 2 more answers

In a titration of 0.35 M HCl and 0.35 M NaOH, how much NaOH should be added to 45.0 ml of HCl to completely neutralize the acid?

Bezzdna [24]

It would be the same amount. So, 45 ml of NaOH is required to be added to the 45 ml of HCI to neutralize the acid fully. Here is a brief calculation:

Firstly, here is your formula: M(HCI) x V(HCI) = M(NaOh) x V(NaOH)
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3 years ago

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3 years ago

Al2(SO4)3(s) + H2O(l) ---- Al2O3(s) + H2SO4 (aq) calculate enthalpy formation for this reaction. Balance the reaction. Calculate

katen-ka-za [31]

The given chemical equation is:

$Al_{2}(SO_{4})_{3}(aq)+H_{2}O(l)-->Al_{2}O_{3}(aq)+H_{2}SO_{4}(aq)$

On balancing the equation we get,

$Al_{2}(SO_{4})_{3}(aq)+3H_{2}O(l)-->Al_{2}O_{3}(aq)+3H_{2}SO_{4}(aq)$

Calculating enthalpy of formation of this reaction from the standard heats of formation of the products and reactants:

Δ $H_{reaction}^{0}=[H_{f}^{0}(Al_{2}O_{3}(s)) + (3*H_{f}^{0}(H_{2}SO_{4}(aq))] - [H_{f}^{0}(Al_{2}SO_{4}(aq)) + (3*H_{f}^{0}(H_{2}O(l))]$

=[(-1669.8kJ/mol)+ {3* (-909.27 kJ/mol)}]-[(-3442kJ/mol)+{3*(-285.8 kJ/mol)}]

=[(-4397.61kJ/mol)]-[(-4299.4kJ/mol)]

=-98.21kJ/mol

Total enthalpy change when 15 mol of $Al_{2}(SO_{4})_{3}$ reacts will be=

$15 mol Al_{2}(SO_{4})_{3}*\frac{-98.21kJ}{1 molAl_{2}(SO_{4})_{3}} =-1473.15kJ/mol$

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4 years ago