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Neko [114]
2 years ago
6

The molar heat capacity of an unknown substance is 92.1 J/mol-K. If the unknown has a molar mass of 118 g/mol, what is the speci

fic heat (J/g-K) of this substance?
Chemistry
1 answer:
dlinn [17]2 years ago
8 0

Answer : The specific heat (J/g-K) of this substance is, 0.780 J/g.K

Explanation :

Molar heat capacity : It is defined as the amount of heat absorbed by one mole of a substance to raise its temperature by one degree Celsius.

1 mole of substance releases heat = 92.1 J/K

As we are given, molar mass of unknown substance is, 118 g/mol that means, the mass of 1 mole of substance is, 118 g.

As, 118 g of substance releases heat = 92.1 J/K

So, 1 g of substance releases heat = \frac{92.1}{118}=0.780J/g.K

Thus, the specific heat (J/g-K) of this substance is, 0.780 J/g.K

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What is the empirical formula? A compound is used to treat iron deficiency in people. It contains 36.76% iron, 21.11% sulfur, an
Mandarinka [93]

Answer: The empirical formula of the compound is Fe_1S_1O_4

Explanation:

Empirical formula is defined formula which is simplest integer ratio of number of atoms of different elements present in the compound.

Percentage of iron in a compound = 36.76 %

Percentage of sulfur in a compound = 21.11 %

Percentage of oxygen in a compound = 42.13 %

Consider in 100 g of the compound:

Mass of iron in 100 g of compound = 36.76 g

Mass of iron in 100 g of compound = 21.11 g

Mass of iron in 100 g of compound = 42.13 g

Now calculate the number of moles each element:

Moles of iron=\frac{36.76 g}{55.84 g/mol}=0.658 mol

Moles of sulfur=\frac{21.11 g}{32.06 g/mol}=0.658 mol

Moles of oxygen=\frac{42.13 g}{16 g/mol}=2.633 mol

Divide the moles of each element by the smallest number of moles to calculated the ratio of the elements to each other

For Iron element = \frac{0.658 mol}{0.658 mol}=1

For sulfur element = \frac{0.658 mol}{0.658 mol}=1

For oxygen element = \frac{2.633 mol}{0.658 mol}=4.001\approx 4

So, the empirical formula of the compound is Fe_1S_1O_4

4 0
1 year ago
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A molecule contains 23.24 g iodine (I),
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Convert each amount of grams into moles:

I: 23.24g x 1 mol / 126.90g = 0.1831 mol I

C: 2.198 x 1 mol / 12.01g = 0.1830 mol C

N: 2.562 x 1 mol / 14.01g = 0.1829 mol N

Each element has roughly the same amount of moles, which means the whole number ratio between the elements is 1:1:1

Therefore the empirical formula is ICN
7 0
1 year ago
What is the pH of a solution made by mixing 15.00 mL of 0.100 M HCl with 50.00 mL of 0.100 M KOH? Assume that the volumes of the
denis23 [38]

Answer:

The correct answer is: pH = 12.73

Explanation:

The <em>neutralization reaction</em> between HCl and KOH is given by the following chemical equation:

HCl + KOH ⇒ KCl + H₂O

Since HCl is a strong acid and KOH is a strong base, HCl is completely dissociated into H⁺ and Cl⁻ ions, whereas KOH is dissociated completely into K⁺ and OH⁻ ions.

For acids, the number of equivalents is given by the moles of H⁺ ions (in this case: 1 equivalent per mol of HCl). For bases, the number of equivalents is given by the moles of OH⁻ ions (in this case: 1 equivalent per mol of KOH).

The H⁺ ions from HCl will react with OH⁻ ions of KOH to give H₂O. The pH is calculated from the difference between the equivalents of H⁺ and OH⁻:

equivalents of H⁺= volume HCl x Molarity HCl

                            = (15.0 mL x 1 L/1000 mL) x 0.100 mol/L

                            = 1.5 x 10⁻³ eq H⁺

equivalents of OH⁻= volume KOH x Molarity KOH

                               = (50.0 mL x 1 L/1000 mL) X 0.100 mol/L

                               = 5 x 10⁻³ eq OH⁻

There are more OH⁻ ions than H⁺ ions. The excess of OH⁻ (that did not react with H⁺ ions) is calculated as follows:

OH⁻ ions= (5 x 10⁻³ eq OH⁻) -  (1.5 x 10⁻³ eq H⁺) = 3.5 x 10⁻³ eq OH⁻= 3.5 x 10⁻³ moles OH⁻  

As the volumes of the solutions are additive, the total volume of the solution is:

V= 15.0 mL + 50.0 mL = 65.0 mL= 0.065 L

So, the concentration of OH⁻ ions in the solution is given by:

[OH⁻] = moles OH⁻/V= (3.5 x 10⁻³ moles OH⁻)/0.065 L = 0.054 mol/L = 0.054 M  

From  [OH⁻], we can calculate pOH:

pOH = -log [OH⁻] = -log (0.054) = 1.27

Finally, we know that pH + pOH= 14; so we calculate pH:

pH= 14 - pOH = 14 - 1,27 =  12.73                                                            

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(hope this helps)
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1 year ago
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