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2 years ago

Xác định nội năng chuẩn ΔU của phản ứng tổng hợp amoniac

Chemistry

1 answer:

Step2247 [10]2 years ago

7 0

Explanation:

xác định nội năng chuẩn ΔU của phản ứng tổng hợp amoniac

ở 400 0 C, biết :

N 2(k) + 3H 2(k) = 2NH 3(k) ΔH 0 T = - 109,0 kJ

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What is the number of moles of potassium chloride present in 148 g of kcl?

OLga [1]

The number of moles of one substance given the amount in mass can be calculated by the use of the molar mass. This is the mass of a compound per 1 mol of the said substance. For, KCl the molar mass is 74.55 g/ mol

148 g / 74.55 g/mol = 2 mol KCl

Hope this answers the question. have a nice day.

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3 years ago

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A potassium ion (K+) would most likely bond with?

ivanzaharov [21]

It's most likely to bond with chlorine.

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3 years ago

An element has two isotopes. One has a mass of 5.03 amu and an abundance of 25%, the other has a mass of 7.20 amu and an

pav-90 [236]

Answer:

Average atomic mass = 6.6575.

Explanation:

As fractions the abundance is 1/4 and 3/4.

The average atomic mass = (5.03 + 3(7.20) / 4

= 6.6575.

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3 years ago

In thermodynamics, we determine the spontaneity of a reaction by the sign of ΔG. In electrochemistry, spontaneity is determined

faltersainse [42]

Answer:

For A: The standard cell potential of the reaction is 4.4 V

For B: The standard Gibbs free energy of the reaction is $-8.50\times 10^5J$

For C: The reaction is spontaneous as written.

Explanation:

For A:

The given chemical reaction follows:

$2Li(s)+Cl_2(g)\rightarrow 2Li^+(aq.)+2Cl^-(aq.)$

The given half reaction follows:

Oxidation half reaction: $Li(s)\rightarrow Li^+(aq.)+e^-;E^o_{Li^+/Li}=-3.04V$ ( × 2)

Reduction half reaction: $Cl_2(g)+2e^-\rightarrow 2Cl^-(aq.);E^o_{Cl_2/2Cl^-}=+1.36V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction.

Here, chlorine will undergo reduction reaction will get reduced.

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o_{cell}=1.36-(-3.04)=4.4V$

Hence, the standard cell potential of the reaction is 4.4 V

For B:

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$

where,

n = number of electrons transferred = $2mol\text{ e}^-$

F = Faradays constant = $96500J/V.mol\text{ e}^-$

$E^o_{cell}$ = standard cell potential = 4.4 V

Putting values in above equation, we get:

$\Delta G^o=-2\times 96500\times 4.4=-849200J=-8.50\times 10^5J$

Hence, the standard Gibbs free energy of the reaction is $-8.50\times 10^5J$

For C:

For a reaction to be spontaneous, the standard Gibbs free energy change of the reaction must be negative.

From above, the standard Gibbs free energy change of the reaction is coming out to be negative.

Hence, the reaction is spontaneous as written.

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4 years ago

Na + H2 → NaCl balanced equation

hodyreva [135]

Answer:

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........

.............

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2 years ago

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