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Arte-miy333 [17]
3 years ago
5

The combustion of 0.0222g of isooctane vapor, C8H18(g), at constant pressure raises the temperature of a calorimeter 0.400 degre

es C. The heat capacity of the calorimeter and water combined is 2.48kJ/degrees C. Find the molar heat of combustion of gaseous isooctane.
Chemistry
1 answer:
saul85 [17]3 years ago
4 0

Answer:

ΔH = 5102.88 KJ/mol

Explanation:

  • Qp = ΔH = CpΔT... constant pressure

∴ Cp = 2.48 KJ/°C

∴ ΔT = 0.400°C

∴ Mw C8H18(g) = 114.22 g/mol

∴ mol C8H18(g) = (0.0222 g)×(mol C8H18/114.22 g) = 1.944 E-4 mol C8H18

⇒ ΔH = (2.48 KJ/°C)×(0.400°C) = 0.992 KJ

⇒ ΔH = 0.992 KJ/ 1.944 E-4 mol C8H18

⇒ ΔH = 5102.88 KJ/mol

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Explanation:

Let us assume that the concentration of [OH^{-} and H^{+} is equal to x. Then expression for K_{w} for the given reaction is as follows.

          K_{w} = [OH^{-}][H^{+}]

          K_{w} = x^{2}

      6.8 \times 10^{-15} = x^{2}

Now, we will take square root on both the sides as follows.

          \sqrt{6.8 \times 10^{-15}} = \sqrt{x^{2}}

          [H^{+}] = 8.2 \times 10^{-8} M

Thus, we can conclude that the H_{3}O^{+} concentration in neutral water at this temperature is 8.2 \times 10^{-8} M.

8 0
3 years ago
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A solution is prepared by dissolving 10.8 g ammonium sulfate in enough water to make 100.0 mL of stock solution. A 10.00- mL sam
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Answer:

Concentration ammonium is 0.036 g/mL

Explanation:

Stock solution has concentration of 0.108g/mL. Take 10mL stock solution and add 50 mL of water meaning it has 1,08g/60mL. Ammonium sulfate has 2 molecules of cation ammonium cation. So, (1,08g/60mL.)*2=0.036 g/mL

8 0
3 years ago
A 6.175 gram sample of an organic compound containing only C, H, and O is analyzed by combustion analysis and 13.30 g CO2 and 5.
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Answer:

Empirical and molecular formulas are the same, C₅H₁₀O₂.

Explanation:

Hello!

In this case, when determining the empirical and molecular formulas of organic compounds via combustion analysis, we first need to compute the moles of carbon and hydrogen via the yielded mass of carbon dioxide and water:

n_C=13.30gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.30molC\\\\n_H=5.447gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O}=0.60molH

Next, we need to compute the mass of oxygen by subtracting the mass of carbon and hydrogen to the mass of the sample of the compound:

m_O =6.175g-0.3molC*12.01gC/molC-0.6molH*1.01gH/molH =1.966gO

And consequently the moles:

n_O=0.12molO

Now, we need to divide the moles of each atom by the fewest moles, it in this case, those of oxygen to obtain the subscripts in the empirical formula:

C=\frac{0.30}{0.12} =2.5\\\\H=\frac{0.60}{0.12} =5\\\\O=\frac{0.12}{0.12} =1

Thus, the empirical formula, taken the nearest whole number is:

C_5H_{10}O_2

Now, if we divide the molar mass of the molecular formula (102.1 g/mol) by that of the empirical formula (102.1 g/mol) we infer they are both the same.

Best regards!

6 0
2 years ago
2. Chromium+oxygen - chromium(III) oxide <br>balanced​
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Answer:

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Explanation:

Information from the question include:

Chromium + oxygen -> chromium(III) oxide

From the word equation given above, the equation can be written as follow:

Cr + O2 —> Cr2O3

The equation can be balance by doing the following:

There are 2 atoms of O on the left side and 3 atoms on the right side. It can be balance by putting 2 in front of Cr2O3 and 3 in front of O2 as shown below:

Cr + 3O2 —> 2Cr2O3

Now, we have 4 atoms of Cr on the right side and 1 atom on the left. It can be balance by putting 4 in front of Cr as shown below:

4Cr + 3O2 —> 2Cr2O3

Now the equation is balanced

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