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3 years ago

Which of the following gives the correct possible values of l for n = 4?

Chemistry

2 answers:

Andreas93 [3]3 years ago

8 0

Answer: The correct answer is the first option. 0, 1, 2, 3

Explanation:

Delicious77 [7]3 years ago

7 0

The answer to your question is a 12

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Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

Now put all the given values in this expression, we get:

$\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]$

$\Delta S^o=-153J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 500 K.

$\Delta G^o=(-1035600J)-(500K\times -153J/K)$

$\Delta G^o=-959100J=-959.1kJ$

Therefore, the value of $\Delta G^o$ for the reaction is -959.1 kJ

3 0

3 years ago

Which statement describes a physical property of copper?

hram777 [196]

Answer:

Copper is typically a solid and has a coppery, bronzy color. It is a metal and has a relatively high melting point. It has a strong luster and can conduct electricity.

7 0

3 years ago

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Calculate the ph of a solution formed by mixing 200.0 ml of 0.30 m hclo with 300.0 ml of 0.20 m kclo. the ka for hclo is 2.9 × 1

masha68 [24]

Answer:

The pH of the solution will be 7.53.

Explanation:

Dissociation constant of KClO= $K_a=2.8\times 10^{-8}$

Concentration of acid in 1 l= 0.30 M

Then in 200 ml = $0.30 M\times 0.200 L=0.06 M$

The concentration of acid, HClO=[acid]= 0.006 M

Concentration of salt in 1 L = 0.20 M

Then in 300 ml = $0.20 M\times 0.300 L=0.06 M$

The concentration of acid, KClO=[salt]= 0.006 M

The pH of the solution will be given by formula :

$pH=pK_{a}^o+\log\frac{[salt]}{[acid]}$

$pH=-\log[2.8\times 10^{-8}]+\frac{[0.06 M]}{[0.06 M]}$

The pH of the solution will be 7.53.

4 0

3 years ago

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Many monatomic ions are found in seawater, including the ions formed from the following list of elements. Write the Lewis symbol

77julia77 [94]

Answer:

The Lewis structures are in image attached.

Explanation:

Lewis symbol is a representation of an element symbol along with its valence electrons around it in the form of dot(s).

Mono-atomic ions formed from the following :

(a) Cl

Chlorine's atomic number is 17 in which only 7 electrons are present in its valence shell .So in order to gain noble gas stability it will gain 1 electron to completes its octet

$Cl=1s^22s^22p^63s^23p^5$

$Cl^-=1s^22s^22p^63s^23p^6$

(b) Na

Sodium's atomic number is 11 in which only 1 electrons are present in its valence shell .So in order to gain noble gas stability it will loose 1 electron to completes its octet. In the Lewis symbol no dot shown as sodium has lost its 1 electron.

$Na=1s^22s^23p^63s^1$

$Na^+=1s^22s^23p^63s^0$

(c) Mg

Magnesium's atomic number is 12 in which only 2 electrons are present in its valence shell .So, in order to gain noble gas stability it will loose 2 electrons to completes its octet.

In the Lewis symbol no dot shown as magnesium has lost its 2 electrons.

$Mg=1s^22s^23p^63s^2$

$Mg^{2+}=1s^22s^23p^63s^0$

(d)Ca

Calcium's atomic number is 20 in which only 2 electrons are present in its valence shell .So, in order to gain noble gas stability it will loose 2 electron to completes its octet.

In the Lewis symbol no dot shown as calcium has lost its 2 electron.

$Ca= 1s^22s^22p^63s^23p^64s^2$

$Ca^{2+}=1s^22s^23p^6^23p^64s^0$

(e) K

Potassium's atomic number is 19 in which only 1 electrons are present in its valence shell .So, in order to gain noble gas stability it will loose 1 electron to completes its octet.

In the Lewis symbol no dot shown as calcium has lost its 1 electron.

$K= 1s^22s^22p^63s^23p^64s^1$

$K^{+}=1s^22s^23p^6^23p^64s^0$

(f) Br

Bromine's atomic number is 35 in which only 7 electrons are present in its valence shell .So in order to gain noble gas stability it will gain 1 electron to completes its octet

$Br=1s^22s^22p^63s^23p^63d^{10}4s^24p^5$

$Br^-= 1s^22s^22p^63s^23p^63d^{10}4s^24p^6$

(g) Sr

Strontium's atomic number is 38 in which only 2 electrons are present in its valence shell .So, in order to gain noble gas stability it will loose 2 electron to completes its octet.

In the Lewis symbol no dot shown as calcium has lost its 2 electron.

$Sr=1s^22s^22p^63s^23p^63d^{10}4s^24p^65s^2$

$Sr^{2+}=1s^22s^22p^63s^23p^63d^{10}4s^24p^65s^0$

(h) F

Florine's atomic number is 7 in which only 7 electrons are present in its valence shell .So in order to gain noble gas stability it will gain 1 electron to completes its octet.

$F=1s^22s^22p^5$

$F^-=1s^22s^22p^6$

5 0

3 years ago

Phosphorus trichloride, PCl3, decomposes to form elemental phosphorus and chlorine. The equation is: 4PCl3 → P4 + 6Cl2. Balance

TiliK225 [7]

19 grams of Cl2 is the correct answer <span>

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5 0

3 years ago

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