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4 years ago

Given the standard entropy values in the table, what is the value of `Delta"Sº"_"reaction"` for the following reaction?

Chemistry

2 answers:

Verizon [17]4 years ago

8 0

Answer:

$\Delta S^o=-149\frac{J}{mol*K}$

Explanation:

Hello,

In this case, we define the standard change in the entropy for a chemical reaction as:

$\Delta S^o=\Sigma \nu _iS_i^o$

Whereas $\nu _i$ is the stoichiometric coefficient of the species i which is negative for products and positive for reactants and $S_i^o$ is the standard entropy for the species i in J/mol*K. Thus, we have:

$\Delta S^o=(-2)(53} )+(-3)(205)+(2)(248)+(2)(38)\\\Delta S^o=-149\frac{J}{mol*K}$

Best regards.

Fantom [35]4 years ago

4 0

Answer:

entropy of reaction = 572-718= -146J/mol-k.

Explanation:

entropy of a reaction = entropy of product - entropy of reactant.

entropy of product = 2×entropy of SO2+2×entropy of NiO

= $2\times 248+2\times 38$

=572 J/mol-k

entropy of reactant = 2×entropy if NiS+3×entropy of O2

= $2\times 53+3\times 205$

=718 J/mol-k

therefore entropy of reaction = 572-718= -146J/mol-k.

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Answer:

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Explanation:

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3 years ago

You use a 15.0 gram piece of aluminum foil to cover a pan in the oven. The specific heat for aluminum is c = 0.900 J/g o C. If t

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Answer:

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Explanation:

Hello,

In this case, we relate the heat, mass, heat capacity and temperature when a thermal change is carried out as shown below:

$Q=mCp(T_{final}-T_{initial})$

Now, for the given data, we compute the absorbed heat (due to the temperature increase) as follows:

$Q=15.0g*0.900\frac{J}{g^oC}*(350^oC-25^oC) \\\\Q=4.39x10^3J=4.39kJ$

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3 years ago

Read 2 more answers

Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 42. g of butane is m

taurus [48]

Answer:

127 grams of carbon dioxide

Explanation:

We need to determine the chemical equation first. Butane has a chemical formula of $C_4H_{10}$ , oxygen is $O_2$ , carbon dioxide is $CO_2$ , and water is $H_2O$ . The reactants are butane and oxygen and the products are carbon dioxide and water. So we write:

$C_4H_{10}+O_2$ ⇒ $CO_2+H_2O$

But remember! We need to balance this. Currently, there are 4 carbon atoms (C), 10 hydrogen atoms (H), and 2 oxygen atoms (O) on the left, while there are 1 carbon atom (C), 2 hydrogen atoms (H), and 3 oxygen atoms (O) on the right. Let's place a coefficient of 4 in front of the carbon dioxide and a coefficient of 5 on the water, so that we have equal numbers of carbon and hydrogen atoms on each side:

$C_4H_{10}+O_2$ ⇒ $4CO_2+5H_2O$

However, we need to ensure that there are equal numbers of O atoms, as well. On the left, we have 2 and on the right we have 13, so let's put a coefficient of 6.5 on the oxygen:

$C_4H_{10}+6.5O_2$ ⇒ $4CO_2+5H_2O$

Finally, multiply everything by 2 to get whole number coefficients:

$2C_4H_{10}+13O_2$ ⇒ $8CO_2+10H_2O$

Ah, now we can actually get to the problem!

We need to determine the limiting reactant, so let's convert the 42 g of butane and 150 g of oxygen into moles of any product, say, carbon dioxide. To convert to moles, we need to find the molar mass of each compound.

The molar mass of butane is 4 * 12.01 + 10 * 1.01 = 58.14 g/mol, while the molar mass of oxygen is 2 * 16 = 32 g/mol. We can now set up the equations:

$42 gC_4H_{10}*\frac{1molC_4H_{10}}{58.14gC_4H_{10}} *\frac{8molCO_2}{2molC_4H_{10}} =2.8896molCO_2$

$150 gO_2*\frac{1molO_2}{32gO_2} *\frac{8molCO_2}{13molO_2} =2.8846molCO_2$

Clearly, we see that 2.8846 < 2.8896, which means that oxygen is the limiting reactant. In other words, the most products can be made when the oxygen is all used up.

Now let's finally convert moles of carbon dioxide into grams by multiplying by its molar mass, which is 12.01 + 2 * 16 = 44.01 g/mol:

$2.8846molCO_2*\frac{44.01gCO_2}{1molCO_2} =127gCO_2$

Notice that we have 3 significant figures because we had 3 significant figures at the start with 150. grams of oxygen.

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3 years ago

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3 years ago

Three different samples were weighed using a different type of balance for each sample. The three were found to have masses of 0

Lena [83]

Answer:

6.096799125kg

Explanation:

According to the question, three different samples weighed using different types of balance had masses: 0.6160959 kg, 3.225 mg, and 5480.7 g.

Based on observation, the mass units in the three measurements are different but must be uniform in order to find the total mass. Hence, we need to convert to the standard unit (S.I unit of mass), which is kilograms (kg)

Since 1kg equals 1,000,000mg

Hence, 3.225mg will be 3.225/1000000

= 0.000003225kg

Also, 1kg equals 1000g

Hence, 5480.7g will be 5480.7/1000

= 5.4087kg

Hence, the total mass of the three samples (now in the same unit) are:

5.4807kg + 0.000003225kg + 0.6160959 kg

= 6.096799125kg

3 0

3 years ago