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3 years ago

Change in enthalpy when 11.2 dm3 of helium at ntp is heated in a cylinder to 100 degrees celsius

Chemistry

1 answer:

REY [17]3 years ago

8 0

Answer:

Enthalpy change = 333.3 J

Explanation:

Given data:

Volume of helium = 11.2 dm³ or 11.2 L

Temperature = 100 °C (100+273= 373 K)

Enthalpy change = ?

Solution:

Normal temperature of room = 20°C (20+273 = 293 K)

one mole of gas occupy volume = 22.4 L

ΔT = final temperature - initial temperature

ΔT = 373K - 293K

ΔT = 80K

Number of moles of gas = 11.2 L/22.4 l

Number of moles of gas = 0.5 mol

Enthalpy change = nRΔT

Enthalpy change = 0.5 mol × 8.3j/mol.k ×80 K

Enthalpy change = 333.3 J

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2 years ago

discuss how variations in electronegativity result in the unequal sharing of electrons in polar molecules.

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Variations in electronegativity prompt in the unequal halves of electrons in polar molecules because when one atom is more electronegative than the other, it becomes more polar than the other.

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1 year ago

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3 years ago

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2 years ago

A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio

Pavlova-9 [17]

Answer:

$C_7H_6O_2$

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

$n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC$

And the moles of hydrogen due to the produced 1.50 grams of water:

$n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O} =0.166molH$

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

$m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO$

Thus, the subscripts in the empirical formula are:

$C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2$

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

$n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol$

Which are equal to the moles of the acid:

$n_{acid}=0.0279mol$

And the molar mass:

$MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol$

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

$C_7H_6O_2$

Best regards!

8 0

2 years ago