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3 years ago

Given the following equation: 8 Fe + S8→ 8 FeS

Chemistry

1 answer:

Mariulka [41]3 years ago

6 0

Answer:

answer the question again and it and change the question ok if less s88 if yes what mass of iron it is needed to wrapped with system that don't deserve that 80 grams of the sulphur age and place with age than that you can you can with

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the atomic mass of magnesium is 24 daltons and the atomic number is 12. a. how many electrons does an electrically neutral magne

vekshin1

Atomic Mass will be 23 the new magnesium formed will be its isotope of magnesium.

We know that,

In stable condition

Number of electrons = Number of protons

Atomic number represents number of proton .

So, here number of proton is 12

Therefore, number of electron is 12

We know that -

Atomic mass = number of protons +Number of neutron

So if magnesium loses one neutron i.e. new number of neutron is 11

then its atomic mass changes.

New atomic mass will be

Atomic mass = $12 + 11$

Atomic mass = 23

This new element with atomic mass 23 and atomic number 12 is the isotope of magnesium.

To know more about isotopes

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3 0

2 years ago

We love blasting our radio in the morning to wake up. Radio waves are used to transmit information on various channels. Calculat

Reil [10]

Answer:

wavelength = 0.56 × 10⁻² m

Explanation:

Given data:

Frequency of radiation = 5.4 × 10¹⁰ Hz

Wavelength = ?

Solution:

Formula:

speed of light = wavelength × frequency

wavelength = speed of light / frequency

wavelength = 3 × 10⁸ m/s / 5.4 × 10¹⁰ Hz ( Hz= 1/s)

wavelength = 0.56 × 10⁻² m

4 0

4 years ago

Which arrow represents the change of state described

Mama L [17]

Answer:N

Explanation:

6 0

3 years ago

. Determine the standard free energy change, ɔ(G p for the formation of S2−(aq) given that the ɔ(G p for Ag+(aq) and Ag2S(s) are

olga nikolaevna [1]

<u>Answer:</u> The standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

<u>Explanation:</u>

We are given:

$K_{sp}\text{ of }Ag_2S=8\times 10^{-51}$

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

R = Gas constant = $8.314J/K mol$

T = temperature = $25^oC=[273+25]K=298K$

K = equilibrium constant or solubility product = $8\times 10^{-51}$

Putting values in above equation, we get:

$\Delta G^o=-(8.314J/K.mol)\times 298K\times \ln (8\times 10^{-51})\\\\\Delta G^o=285793.9J/mol=285.794kJ$

For the given chemical equation:

$Ag_2S(s)\rightleftharpoons 2Ag^+(aq.)+S^{2-}(aq.)$

The equation used to calculate Gibbs free change is of a reaction is:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]$

The equation for the Gibbs free energy change of the above reaction is:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(Ag^+(aq.))})+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times \Delta G^o_f_{(Ag_2S(s))})]$

We are given:

$\Delta G^o_f_{(Ag_2S(s))}=-39.5kJ/mol\\\Delta G^o_f_{(Ag^+(aq.))}=77.1kJ/mol\\\Delta G^o=285.794kJ$

Putting values in above equation, we get:

$285.794=[(2\times 77.1)+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times (-39.5))]\\\\\Delta G^o_f_{(S^{2-}(aq.))=92.094J/mol$

Hence, the standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

8 0

4 years ago

For the reaction at 298 K, 2NO2(g) N2O4(g) the values of ΔH° and ΔS° are -58.03 kJ and -176.6 J/K, respectively. Calculate the v

Ierofanga [76]

Answer:

$\Delta G^o=-5.4032 kJ$

The temperature for $\Delta G^o=0[/tex is [tex]T=328.6 K$

Explanation:

The three thermodinamic properties (enthalpy, entropy and Gibbs's energy) are linked in the following formula:

$\Delta G^o=\Delta H^o + T*\Delta S^o$

Where:

$\Delta G^o$ is Gibbs's energy in kJ

$\Delta H^o$ is the enthalpy in kJ

$\Delta S^o$ is the entropy in kJ/K

$T$ is the temperature in K

Solving:

$\Delta G^o=-58.03 kJ - 298K*-0.1766 kJ/K$

$\Delta G^o=-5.4032 kJ$

For $\Delta G^o=0$ :

$0=\Delta H^o - T*\Delta S^o$

$\Delta H^o= T*\Delta S^o$

$T=\frac{\Delta H^o}{\Delta S^o}$

$T=\frac{-58.03 kJ}{-0.1766 kJ/K}$

$T=328.6 K$

3 0

3 years ago

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