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MrMuchimi
3 years ago
7

Humans started using elemental copper about 6000 years ago and started using elemental tin about 3800 years ago. Use the heat of

formation values for copper C (II) oxide and tin (IV) oxide to help explain why humans were able to use elemental copper before they were able to use elemental tin.
Chemistry
1 answer:
Elis [28]3 years ago
5 0

Answer:

As the Tin has "lousy properties," humans can use copper before tin.

Explanation:

Heat of formation of the copper (II) oxide = -156KJ/mol

Heat of formation of Tin (IV) oxide =-581KJ/mol

As, we know that the copper will form the copper oxide after receiving reduced energy. Humans were able to use the elemental copper before tin. And also, tin has itself the lousy properties as compared to Copper. Copper are also the good conductor of heat, hence they are used in the electric wiring. Apart, from that it is also being used for roofing and plumbing. And as heat exchangers in the industries Copper also possess the antibacterial property.

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Answer:

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Explanation:

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Which is a physical property of matter?
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What is 4,678,679 dL in scientific notation
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8 0
4 years ago
How many grams of water can be formed when 80 grams of sodium hydroxide (NaOH) reacts with an excess of sulfuric acid, H2SO4, in
podryga [215]

Answer:

36g of H2O.

Explanation:

We'll begin by writing the balanced equation for the reaction:

2NaOH + H2SO4 —> Na2SO4 + 2H2O

Next, we shall determine the mass of NaOH that reacted and the mass of H2O produced from the balanced equation. This is illustrated below:

Molar mass of NaOH = 23 + 16 + 1 = 40g/mol

Mass of NaOH from the balanced equation = 2 x 40 = 80g

Molar mass of H2O = (2x1) + 16 = 18g/mol

Mass of H2O from the balanced equation = 2 x 18 = 36g.

From the balanced equation above, we can see evidently that:

80g of NaOH reacted to produce 36g of H2O.

5 0
4 years ago
Pentaborane-9, B5H9, is a colorless, highly reactive liquid that will burst into flame when exposed to oxygen. The reaction is 2
mina [271]

<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

8 0
3 years ago
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