7) Which element would have the lowest first ionization energy? * Li Be Na Mg

Compound A has a partition coefficient (K) of 7 when comparing its solubility in CH2Cl2 to water ( K=11, [solubility of A in g/m

Hitman42 [59]

Answer:

The equation which will tell us how much of A that is mis inmthe water layer after partitioning is: 7 = (17 - x) g / 150 mL ÷ x g /100 mL

Explanation:

A partition coefficient is the ratio of the concentration of a substance in one solvent phase to the concentration in a second solvent phase when the two concentrations are at equilibrium. Usually the two phases are an organic phase and an aqueous phase. Thus, the partition coefficient K, of a compound is the ratio of the compound's concentration in the organic layer compared to the aqueous layer.

K = C₁/C₂ at equilibrium

In the compound A given, CH₂Cl₂ is the organic phase while water is the aqueous phase

Amount of A that is partitioned in between dichloromethane, CH₂Cl₂ and water, H₂O is 17.0 g

Let the amount of A that is dissolved in water be x g

Solubility of A in water given in g/mL = (x / 100) g/ml

Amount of A dissolved in dichloromethane, CH₂Cl₂ = (17 - x) g

Solubility of A in dichloromethane, CH₂Cl₂ given in g/mL = (17 - x/150) g/mL

Since the partition coefficient, K of compound A when comparing its solubility in CH₂Cl₂ to water is 7, that is;

K = [solubility of A in g/ml in CH₂Cl₂] / [solubility of A in g/ml in H2O] = 7

The equation for the amount of A in the water layer is given as follows:

7 = (17 - x) g / 150 mL ÷ x g /100 mL

Solving for x

7 = (17 -x) × 100 / 150x

7 × 150x = (17 - x) × 100

1050x = 1700 - 100x

1150x = 1700

x = 1700/1150

x = 1.48 g

5 0

3 years ago

Menthol, the substance we can smell in mentholated cough drops, is composed of C, H, and O. A 0.1005 g sample of menthol is comb

Bogdan [553]

Answer: The molecular formula for the menthol is $C_{10}H_{20}O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=0.2829g$

Mass of $H_2O=0.1159g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.2829 g of carbon dioxide, $\frac{12}{44}\times 0.2829=0.077g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 0.1159 g of water, $\frac{2}{18}\times 0.1159=0.0129g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.1005) - (0.077 + 0.0129) = 0.0106 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.077g}{12g/mole}=0.0064moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.0129g}{1g/mole}=0.0129moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.0106g}{16g/mole}=0.00066moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00066 moles.

For Carbon = $\frac{0.0064}{0.00066}=9.69\approx 10$

For Hydrogen = $\frac{0.0129}{0.00064}=19.54\approx 20$

For Oxygen = $\frac{0.00066}{0.00066}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 10 : 20 : 1

Hence, the empirical formula for the given compound is $C_{10}H_{20}O_1=C_{10}H_{20}O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 156.27 g/mol

Mass of empirical formula = 156 g/mol

Putting values in above equation, we get:

$n=\frac{156.27g/mol}{156g/mol}=1$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 10)}H_{(1\times 20)}O_{(1\times 1)}=C_{10}H_{20}O$

Thus, the molecular formula for the menthol is $C_{10}H_{20}O$

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