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2 years ago

Why does 5060 have three significant figures?

Chemistry

1 answer:

Morgarella [4.7K]2 years ago

3 0

<h2>5060 have three significant figures : Explanation given below </h2>

Explanation:

Significant figures

The significant figures (also known as the significant digits and decimal places) of a number are digits that possess certain meaning .

It includes all digits except: zeros

Rules to find significant figures

1.All non-zero digits are considered significant. For example, 23 has two significant figures.

2.Zeros in between two non-zero digits are significant: like in 202.1201 has seven significant figures.

3.Zeros to the left of the significant figures are not significant. For example, .000021 has two significant figures, zeros have no value .

4.Zeros to the right of the significant figures are significant.

That is the reason in number 5060 , it has 3 significant figures .

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Hydrogen, a potential future fuel, can be produced from carbon (from coal) and steam by the following reaction: C(s)+2H2O(g)→2H2

madam [21]

The question is incomplete , complete question is:

Hydrogen, a potential future fuel, can be produced from carbon (from coal) and steam by the following reaction:

$C(s)+ 2 H_2O(g)\rightarrow 2H_2(g)+CO_2(g).\Delta H=?$

Note that the average bond energy for the breaking of a bond in CO2 is 799 kJ/mol. Use average bond energies to calculate ΔH of reaction for this reaction.

Answer:

The ΔH of the reaction is -626 kJ/mol.

Explanation:

$C(s)+ 2 H_2O(g)\rightarrow 2H_2(g)+CO_2(g).\Delta H=?$

We are given with:

$\Delta H_{H-O}=459 kJ/mol$

$\Delta H_{H-H}=432 kJ/mol$

$\Delta H_{C=O}=799 kJ/mol$

ΔH = (Energies required to break bonds on reactant side) - (Energies released on formation of bonds on product side)

$\Delta H=(4\times \Delta H_{O-H})-(2\times \Delta H_{H-H}+2\times\Delta H_{C=O})$

$=(4\times 459 kJ/mol)-(2\times 432 kJ/mol+2\times 799 kJ/mol$

$\Delta H=-626 kJ/mol$

The ΔH of the reaction is -626 kJ/mol.

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