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kvasek [131]
3 years ago
6

Plz hurry fast I need help !!!

Chemistry
2 answers:
NikAS [45]3 years ago
7 0

The Second one is Correct

This is   because there three extra protons (+) than electrons

Vikki [24]3 years ago
5 0

Answer:

USE SOCRATIC

Explanation:

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A binary compound of oxygen with another element is called oxide. An oxide is a binary compound of oxygen and another element. Oxygen combines with metals and non-metals to form respective oxides.

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In atmospheric chemistry, the following chemical reaction converts SO2, the predominant oxide of sulfur that comes from combusti
Misha Larkins [42]

Answer:

Explanation:

From the given information;

The chemical reaction can be well presented as follows:

\mathtt{SO_{2(g)} + \dfrac{1}{2}O_{2(g)} }  ⇄ \mathtt{3SO_{2(l)}}

Now, K is known to be the equilibrium constant and it can be represented in terms of each constituent activity:

i.e

K = \dfrac{a_{so_3}}{a_{so_2} a_{o_2}^{\frac{1}{2}}}

However, since we are dealing with liquids solutions;

K = \dfrac{1}{\dfrac{Pso_2}{P^0}\Big ( \dfrac{Po_2}{P^0} \Big)^{1/2}}   since the activity of a_{so_3} is equivalent to 1

Hence, under standard conditions(i.e at a pressure of 1 bar)

K = \dfrac{1}{Pso_2Po_2^{1/2}}

(b)

From the CRC Handbook, we are meant to determine the value of the Gibb free energy by applying the formula:

\Delta _{rxn} G^o = \sum \Delta_f \ G^o (products) - \sum \Delta_fG^o (reactants) \\ \\ = (1) (-368 \ kJ/mol) - (\dfrac{1}{2}) (0) - ((1) (-300.13 \ kJ/mol)) \\ \\ = -368 \ kJ/mol + 300.13 \ kJ/mol \\ \\  \simeq -68 \ kJ/mol

Thus, for this reaction; the Gibbs frree energy = -68 kJ/mol

(c)

Le's recall that:

At equilibrium, the instantaneous free energy is usually zero &

Q(reaction quotient) is equivalent to K(equilibrium constant)

So;

\mathtt{\Delta _{rxn} G = \Delta _{rxn} G^o + RT In Q}

\mathtt{0- \Delta _{rxn} G^o = RTIn K } \\ \\ \mathtt{ \Delta _{rxn} G^o = -RTIn K }  \\ \\  K = e^{\dfrac{\Delta_{rxn} G^o}{RT}} \\ \\  K = e^{^{\dfrac{67900 \ J/mol}{8.314 \ J/mol \times 298 \ K}} }

K =7.98390356\times 10^{11} \\ \\  \mathbf{K = 7.98 \times 10^{11}}

(d)

The direction by which the reaction will proceed can be determined if we can know the value of Q(reaction quotient).

This is because;

If  Q < K, then the reaction will proceed in the right direction towards the products.

However, if Q > K , then the reaction goes to the left direction. i.e to the reactants.

So;

Q= \dfrac{1}{Pso_2Po_2^{1/2}}

Since we are dealing with liquids;

Q= \dfrac{1}{1 \times 1^{1/2}}

Q = 1

Since Q < K; Then, the reaction proceeds in the right direction.

Hence, SO2 as well O2 will combine to yield SO3, then condensation will take place to form liquid.

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3 years ago
You estimate that you sent 135 texts last week but when you get the bill, it was actually 175. Calculate the percent error
FinnZ [79.3K]

Answer: -40

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What is a cercaria?<br> *zoology
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Type the correct answer in the box. sodium carbonate (na2co3) is sometimes used as a water-softening agent. suppose that a worke
Natali [406]

The mass of dissolved sodium carbonate is 109 g of Na₂CO₃.

<h3>What is a solution ?</h3>

A solution consists of two elements , a solute and a solvent .

A solute is dissolved into solvent to make a solution.

The concentration of a solution is measure by Molarity , Molality , mass% etc.

It is given in the question that

sodium carbonate ( Na₂CO₃) is sometimes used as a water-softening agent

a worker prepares a 0.724 m solution of  Na₂CO₃ and water.

the volume of the solution is 1.421 liters.

Mass of dissolved sodium carbonate is grams =?

Concentration = 0.724 M

Volume = 1.421 liters

Molecular mass of  Na₂CO₃  = 106 g

The mass of the solute needs to be determined

Molarity = moles / volume\

moles = Molarity x volume

Substituting the values

moles = 0.724 x 1.421

moles = 1.029

1 mol of Na₂CO₃ means  106 gm of  Na₂CO₃

Then 1.029 moles will mean 1.029 * 106

109.07 gm

Therefore , the mass of dissolved sodium carbonate is 109 grams.

To know more about Solution

brainly.com/question/1616939

#SPJ1

         

7 0
2 years ago
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