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3 years ago

How do ions, anions, and cations bond to create electronic neutrality?

1 answer:

5 0

Every atom in its ground state is uncharged. It has according to its atomic number, the same number of protons and electrons. Electrons are rather labile, however, and an atom will often gain or lose them depending on its electronegativity.

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2. In the reaction NO + NO2 ⇌ N2O3, an experiment finds equilibrium concentrations of [NO] = 3.8 M, [NO2] = 3.9 M, and [N2O3] =

notsponge [240]

Kc= concentration of product divided by concentration of reactant
NO + NO2 ----> N2O3

Kc =(N2O3) / (No)(NO2)

Kc= ( 1.3 )/{ (3.9)(3.8) }

Kc=0.088 ( answer B)

4 0

3 years ago

Read 2 more answers

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

A thief who has stolen large quantities of uranium-235 may have increased

Gala2k [10]

Nuclear waste not a b or d it’s nuclear waste

3 0

3 years ago

A student measures the mass of a sample as 9.64 g. Calculate the percentage error, given that the correct mass is 9.80 g.

rosijanka [135]

Answer:

<h3>The answer is 1.63 %</h3>

Explanation:

The percentage error of a certain measurement can be found by using the formula

$P(\%) = \frac{error}{actual \: \: number} \times 100\% \\$

From the question

actual mass = 9.80 g

error = 9.80 - 9.64 = 0.16

We have

$p(\%) = \frac{0.16}{9.80} \times 100 \\ = 1.632653061...$

We have the final answer as

Hope this helps you

5 0

4 years ago

For the following experiments, identify the independent variable and the dependent variable:

faust18 [17]

Answer:

A) IV: number of hours DV: score

Explanation:

Coz,

IV- is the variable that u change

DV- is the variable that u test or measure

In this case, we need to change the hours of studying to see how studying for different hours [1 hr, 2hr, etc (( Independant variable coz we're changing the no. of hours)) ] is going to affect the student's score, and in the end, we are measuring the score of each student [this is the dependant variable].

3 0

3 years ago