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marshall27 [118]
4 years ago
8

Can a substance be cooled to a temperature below its freezing point

Chemistry
1 answer:
Fittoniya [83]4 years ago
5 0
Pure, crystalline solids have a characteristic melting point, the temperature at which the solid melts to become a liquid. The transition between the solid and the liquid is so sharp for small samples of a pure substance that melting points can be measured to 0.1oC. The melting point of solid oxygen, for example, is -218.4o<span>C.</span>
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Using the equation 2H2+O2 = 2H2I, if 192 g if oxygen reacts completely, how many grams of hydrogen must react with it
dimaraw [331]
I assume there is a typo in the equation.  It is H2O instead of H2I.

The ratio of H2 and O2 that react with each other is 2:1.  To find out the grams of H2, we need to first find out the moles of O2 and H2.

Moles of O2 = mass of O2/molar mass of O2 = 192g/32g/mol = 6 mol.  Therefore, the moles of H2 = 6mol *2 = 12 mol.

So the mass of H2 that reacts with O2 = moles of H2 * molar mass of H2 = 12 mol * 2 g/mol = 24 g

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Iron plus what equals iron chloride
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For a particular isomer of C8H18, the combustion reaction produces 5113.3 kJ of heat per mole of C8H18(g) consumed, under standa
mart [117]

Answer: The standard enthalpy of formation of this isomer of C_8H_{18}(g) is -210.9 kJ

Explanation:

The given balanced chemical reaction is,

C_8H_{18}(g)+\frac{25}{2}O_2(g)\rightarrow 8CO_2(g)+9H_2O(g)

First we have to calculate the enthalpy of formation of  C_8H_{18}.

\Delta H^o=H_f_{product}-H_f_{reactant}

\Delta H^o=[n_{CO_2}\times \Delta H_f^0_{(CO_2)}+n_{H_2O}\times \Delta H_f^0_{(H_2O)}]-[n_{O_2}\times \Delta H_f^0_{(O_2)+n_{C_8H_{18}}\times \Delta H_f^0_{(C_8H_{18})}]

where,

We are given:

\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.8kJ/mol\\\Delta H^o_f_{(O_2(g)))}=0kJ/mol

Putting values in above equation, we get:

-511.3kJ/mol=[(8\times -393.5)+(9\times -241.8)]-[(\frac{25}{2}\times 0)+(1\times \Delta H_f^0_{(C_8H_{18})}

\Delta H_f^0_{(C_8H_{18})}=-210.9kJ

7 0
4 years ago
Read 2 more answers
When do you use 6.02*10^23?? Do you use it when going to atoms or molecules????
Andrew [12]
You would use it in both atoms and molecules if it’s in are large quantity
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4 years ago
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