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ollegr [7]
3 years ago
5

A 1.450 g sample of an unknown organic compound , X, is dissolved in 15.0 g of toluene

Chemistry
1 answer:
muminat3 years ago
4 0

Answer:

Molecular weight of the compound = 372.13 g/mol

Explanation:

Depression in freezing point is related with molality of the solution as:

\Delta T_f = K_f \times m

Where,

\Delta T_f = Depression in freezing point

K_f = Molal depression constant

m = Molality

\Delta T_f = K_f \times m

1.33 = 5.12 \times m

m = 0.26

Molality = \frac{Moles\ of\ solute}{Mass\ of\ solvent\ in\ kg}

Mass of solvent (toluene) = 15.0 g = 0.015 kg

0.26 = \frac{Mole\ of\ compound}{0.015}

Moles of compound = 0.015 × 0.26 = 0.00389 mol

Mol = \frac{Mass\ in\ g}{Molecular\ weight}

Mass of the compound = 1.450 g

Molecular\ weight = \frac{Mass\ in\ g}{Moles}

Molecular weight = \frac{1.450}{0.00389} = 372.13\ g/mol

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Answer:

There are 4 tryptophans in the protein.

Explanation:

According to question,  protein contains one tyrosine residue and say x number of tryptophans.

Concentration of protein solution = 1.0 micromolar = 1.0\times 10^{-6} Molar

Molar absorptivity of a protein solution : \epsilon

\epsilon = \epsilon _{tyro}+\epsilon _{tryp}

=1\times 2000 M^{-1}cm^{-1}+x\times 5500 M^{-1}cm^{-1}

Length of the cuvette = l = 1.0 cm

Absorbance of protein solution at 280 nm = A = 0.024

A=\epsilon \times l\times c ( Beer-Lambert's law)

0.024=(1\times 2000 M^{-1}cm^{-1}+x\times 5500 M^{-1}cm^{-1})\times 1 cm\times 1.0\times 10^{-6} M

Solving for x :

x = 4

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7 0
3 years ago
How many grams of calcium chloride will be produced when 29.0 g of calcium carbonate is combined with 15.0 g of hydrochloric aci
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Answer:

22.7 g of CaCl₂ are produced in the reaction

Explanation:

This is the reaction:

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Now, let's determine the limiting reactant.

Let's divide the mass between the molar mass, to find out moles of each reactant.

29 g / 100.08 g/m = 0.289 of carbonate

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1 mol of carbonate must react with 2 moles of acid

0.289 moles of carbonate will react with the double of moles (0.578)

I only have 0.411 of HCl, so the acid is the limiting reactant.

Ratio is 2:1, so I will produce the half of moles, of salt.

0.411 / 2 = 0.205 moles of CaCl₂

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lara31 [8.8K]

Answer:

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Explanation:

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Now from the chemical reaction we devise the following reasoning:

if         3 moles of NH₃ are produce 1 mole of (NH₄)₃PO₄

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mass = 0.103 × 149 = 15.35 g of (NH₄)₃PO₄

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