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2 years ago

Calculate the number of molecules in 6.00 moles h2s?.

2 answers:

6 0

Use the equation: Nm = M x Av
Nm = Moles x Avogadro constant
Nm = 6 x 6.02x10²³
Nm = 3.612x10²⁴

so in six moles of H2S there are 3.612x10²⁴ molecules

hope that helps

LenKa [72]2 years ago

5 0

To calculate the number of molecules in 6.00 moles of hydrogen sulfide, H2S, the equivalence factor used is Avogadro's number equal to (6.022 x 10^23 molecules/mole). The answer is 6.00 moles of hydrogen sulfide * (6.022 x 10^23 molecules/mole) equal to 3.61 x10^24 molecules.

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2 years ago

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kari74 [83]

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2 years ago

Read 2 more answers

Under standard-state conditions, which of the following species is the best reducing agent? a. Ag+ b. Pb c. H2 d. Ag e. Mg2+

eimsori [14]

Answer: The correct answer is Option b.

Explanation:

Reducing agents are defined as the agents which help the other substance to get reduced and itself gets oxidized. They undergo oxidation reaction.

$X\rightarrow X^{n+}+ne^-$

For determination of reducing agents, we will look at the oxidation potentials of the substance. Oxidation potentials can be determined by reversing the standard reduction potentials.

For the given options:

Option a: $Ag^+$

This ion cannot be further oxidized because +1 is the most stable oxidation state of silver.

Option b: $Pb$

This metal can easily get oxidized to $Pb^{2+}$ ion and the standard oxidation potential for this is 0.13 V

$Pb\rightarrow Pb^{2+}+2e^-;E^o_{(Pb/Pb^{2+})}=+0.13V$

Option c: $H_2$

This metal can easily get oxidized to $H^{+}$ ion and the standard oxidation potential for this is 0.0 V

$H_2\rightarrow 2H^++2e^-;E^o_{(H_2/H^{+})}=0.0V$

Option d: $Ag$

This metal can easily get oxidized to $Ag^{+}$ ion and the standard oxidation potential for this is -0.80 V

$Ag\rightarrow Ag^{+}+e^-;E^o_{(Ag/Ag^{+})}=-0.80V$

Option e: $Mg^{2+}$

This ion cannot be further oxidized because +2 is the most stable oxidation state of magnesium.

By looking at the standard oxidation potential of the substances, the substance having highest positive $E^o$ potential will always get oxidized and will undergo oxidation reaction. Thus, considered as strong reducing agent.

From the above values, the correct answer is Option b.

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2 years ago