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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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4 years ago

You have two sealed jars of water at the same temperature. In the first jar there is a large amount of water. In the second jar

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since both the jars are kept at the same temperature the vapor pressure will be same in both the cases.

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3 years ago

When 0.42 g of a compound containing C, H, and O is burned completely, the products are 1.03 g CO2 and 0.14 g H2O. The molecular

Luda [366]

Answer: The empirical and molecular formula for the given organic compound is $C_2H_2O_4$ and $C_6H_4O_2$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=1.03g$

Mass of $H_2O=0.14g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 1.03 g of carbon dioxide, $\frac{12}{44}\times 1.03=0.28g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 0.14 g of water, $\frac{2}{18}\times 0.14=0.016g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.42) - (0.28 + 0.016) = 0.124 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.28g}{12g/mole}=0.023moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.016g}{1g/mole}=0.016moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.124g}{16g/mole}=0.00775moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00775 moles.

For Carbon = $\frac{0.023}{0.00775}=2.96\approx 3$

For Hydrogen = $\frac{0.016}{0.00775}=2.06\approx 2$

For Oxygen = $\frac{0.00775}{0.00775}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 2 : 1

Hence, the empirical formula for the given compound is $C_3H_{2}O_1=C_3H_2O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 108.10 g/mol

Mass of empirical formula = 54 g/mol

Putting values in above equation, we get:

$n=\frac{108.10g/mol}{54g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(3\times 2)}H_{(2\times 2)}O_{(1\times 2)}=C_6H_4O_2$

Thus, the empirical and molecular formula for the given organic compound is $C_3H_2O$ and $C_6H_4O_2$

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3 years ago

The painkiller, Advil® contains the active ingredient ibuprofen (IB), which has a pKb of

denis23 [38]

This problem is providing the basic dissociation constant of ibuprofen (IB) as 5.20, its pH as 8.20 and is requiring the equilibrium concentration of the aforementioned drug by giving the chemical equation at equilibrium it takes place. The obtained result turned out to be D) 4.0 × 10−7 M, according to the following work:

First of all, we set up an equilibrium expression for the given chemical equation at equilibrium, in which water is omitted for it is liquid and just aqueous species are allowed to be included:

$Kb=\frac{[IBH^+][OH^-]}{[IB]}$

Next, we calculate the concentration of hydroxide ions and the Kb due to the fact that both the pH and pKb were given:

$pOH=14-8.20=5.80$

$[OH^-]=10^{-5.8}=1.585x10^{-6}M$

$Kb=10^{-5.20}=6.31x10^{-6}$

Then, since the concentration of these ions equal that of the conjugated acid of the ibuprofen (IBH⁺), we can plug in these and the Kb to obtain:

$6.31x10^{-6}=\frac{(1.585x10^{-6})(1.585x10^{-6})}{[IB]}$

Finally, we solve for the equilibrium concentration of ibuprofen:

$[IB]=\frac{(1.585x10^{-6})(1.585x10^{-6})}{6.31x10^{-6}}=4.0x10^{-7}$

Learn more:

(Weak base equilibrium calculation) brainly.com/question/9426156

4 0

3 years ago

Give the formula for an ionic compound formed from each pair of ions

Vera_Pavlovna [14]

Please, find the complete question in the picture attached.

Answer:

1. Na₂O

2. AlF₃

3. MgO

4. Ca₃P₂

Explanation:

1. Na⁺ and O²⁻

An ionic compound will be formed when two ions of opposite charge attract each other.

The positive ion is called cation and the negative ion is called anion.

Thus, every ionic compound has a cation and an anion electrostatically bonded.

The ions must combine in a proportion that genders a neutral compound: so you must have as many cations as negative charge has one anion and as many anions as positive charge has one cation.

This is, if the cation has charge +x, there will be x anions, and if the anion has charge -y, there will be y cations.

Simbolically:

$Cation:A^{+x}\\ \\ Anion:B^{-y}\\ \\ Compound:A_yB_x$

As you see, the subscripts for each element in the chemical formula are obtained by the exchage of the charges of the ions.

Then, for Na⁺ and O²⁻, the subscript for Na will be 2 and the subscript for O will be 1; and the formula of the ionic compound formed by this pair ot ions is:

Na₂O

2. Al³⁺ and F⁻

The subscript of Al will be 1 (because the F ion has charge 1-, and the subscript of F will be 3 (because the Al ion has charge +3).

Thus the ionic compound formed by this pair of ions is:

AlF₃

3. Mg²⁺ and S²⁻

The charge 2+ from Mg atom wil become the subscript 2 of S atom, and the charge 2- will become the subscript 2 of Mg atom:

That results in the formula: Mg₂S₂

Except for some special compounds, the chemical formula is simplified, dividing by the least common denominator. In this case, that means that the two 2 subscripts are simplified to 1, and the final chemical formula for the ionic compound formed by this pair of ions is:

4. Ca²⁺ and P³⁻

The charge 2+ from Ca will become subscript 2 for P and the charge 3- whill become subscript 3 for Ca.

Hence, the ionic compound formed by these two ions is:

Ca₃P₂

7 0

3 years ago