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3 years ago

How does the number of coupons you use affect the amount of money you save?

1 answer:

7 0

Using coupons seems like an old-fashioned, time-consuming way to save a few cents on the products you buy at the grocery store. After all, most coupons don't usually have a face value of more than $1 and, quite often, they're worth far less. But done properly, couponing can save you thousands of dollars every year.

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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

What quantity of heat is transferred when a 150.0g block of iron metal is heated from 25.0°C to 73.3°C? What is the direction of

s344n2d4d5 [400]

The direction of heat flow is increased which means blocks temperature is higher and hotter than it was before

3 0

2 years ago

1. What is the potential energy of a 8 kg ball that is at a height of 20 m?

weqwewe [10]

Answer:

1. 1568 J

2. 0 J

3. 1176 J

Explanation:

PE = mgh

(PE = Potential Energy) = (m = mass)(g = gravitational force which is 9.8)(h = height)

1. (3)(9.8)(20) = 1568 J

2. PE = (3)(9.8)(0) = 0 J

3. (5)(9.8)(24) = 1176 J

4 0

2 years ago

⇔what process occurs in some cells

Serhud [2]

It could be a couple different thing, explain more.

3 0

2 years ago

A student pours 44.3 g of water at 10Â°C into a beaker containing 115.2 g of water at 10Â°C. What are the final mass, temperatur

Agata [3.3K]

Answer:

Final mass = 159.5 g

Final temperature = 10 C

Final density = 1.00 g/ml

Explanation:

Given:

Beaker 1:

Mass of water = 44.3 g

Temperature = 10 C

Beaker 2:

Mass of water = 115.2 g

Temperature = 10 C

Density of water at 10C = 1.00 g/ml

To determine:

The final mass, temperature and density of water

Calculation:

$Final\ mass\ of \ water = Beaker\ 1 + Beaker\ 2 = 44.3 + 115.2 = 159.5 g$

Since there is no change in temperature, the final temperature will be 10 C

Density of a substance is an intensive property i.e. it is independent of the mass. Hence the density of water will remain constant i.e. 1.00 g/ml

3 0

3 years ago