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3 years ago

If 11.2 g of naphthalene, C10H8, is dissolved in 107.8 g of chloroform, CHCl3, what is the molality of the solution?

Chemistry

1 answer:

Art [367]3 years ago

5 0

Answer: 0.811m

Explanation: Molarity is the moles of solute present per kg of solvent.

Molar mass of C10H8 = . 128.17 g/mol

No. of moles = given mass/molar mass

11.2g C10H8 = 11.2/128.17 g/mol C10H8

= 0.0874 mol C10H8

NOW 107.8g solvent (chloroform) contains 0.0879 molecules (naphthalene)

1000g '' will contain (0.0879/107.8 x 1000) mole solute

= 0.811 mole

Molarity of the solution is 0.811 m (option A)

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Which one of the following compounds will be least soluble in water?

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Water is a polar solvent. Since, 'Like dissolves like' a polar compound will be most soluble in water whereas a non-polar compound will be the least soluble.

The increasing order of polarity for the given organic compounds is:-

CH3OCH3 < CH3NH2 < CH3CHO < CH3CH2OH

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2 years ago

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A monatomic ideal gas that is initially at 1.50 * 105 Pa and has a volume of 0.0800 m3 is compressed adiabatically to a volume o

lubasha [3.4K]

Answer:

300000Pa or 3×10^5 Pa

Explanation:

Since the problem involves only two parameters of volume and pressure, the formula for Boyle's law is suitably used.

Using Boyle's law

P1V1 = P2V2

P1 is the initial pressure = 1.5×10^5Pa

V1 is the initial volume = 0.08m3

P2 is the final pressure (required)

V2 is the final volume = 0.04 m3

From the formula, P2 = P1V1/V2

P2 = 1.5×10^5 × 0.08 ÷ 0.04

= 300000Pa or 3×10^5 Pa.

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3 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!

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3 years ago

The atoms of a solid aluminum can are close together, vibrating in a rigid structure. If the can is warmed up on a hot plate, wh

Oksana_A [137]

Explanation:

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As a result, when can is warmed up then atoms of solid aluminium can will start to vibrate more because heat will decrease the force of attraction between the molecules.

Thus, atoms will start to move away from each other and hence, they will vibrate more.

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2 years ago

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A student performs an experiment similar to what you will be doing in lab, except using titanium metal instead of magnesium meta

Genrish500 [490]

Answer:

n = 0.0022 mol

Explanation:

Moles is denoted by given mass divided by the molar mass ,

Hence ,

n = w / m

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w = 0.108 g

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3 years ago