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20 days ago

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Which of the following solutions will have the lowest freezing point? Input the appropriate letter. A. 35.0 g of C3H8O in 250.0

g of ethanol (C2H5OH) B. 35.0 g of C4H10O in 250.0 g of ethanol (C2H5OH) C. 35.0 g of C2H6O2 in 250.0 g of ethanol (C2H5OH)

1 answer:

VMariaS [2.9K]20 days ago

4 0

Answer:

The solution composed of 35.0 g of $C_3H_8O$ mixed with 250.0 g of ethanol will exhibit the lowest freezing temperature

Explanation:

$\Delta T_f=K_f\times m$

where,

$\Delta T_f$ = reduction in freezing point =

$K_f$ = freezing point constant

m = molality

As observed, a higher molality in the solution corresponds to a greater depression in the freezing point, resulting in a more reduced freezing point of the solution.

$Molality=\frac{moles}{\text{mass of solvent in kg}}$

A. 35.0 g of $C_3H_8O$ in 250.0 g of ethanol.

The amount of moles of $C_3H_8O$ = $\frac{35.0 g}{60 g/mol}=0.5833 mol$

The mass of ethanol solvent is 250.0 g = 0.25 kg (1 g = 0.001 kg)

$m=\frac{0.5833 mol}{0.25 kg}=2.33 m$

B. 35.0 g of $C4H_{10}O$ in 250.0 g of ethanol

Moles of $C_4H_{10}O$ $=[tex]\frac{35.0 g}{74 g/mol}=0.4730 mol$

The mass of ethanol solvent is 250.0 g = 0.25 kg (1 g = 0.001 kg)

$m'=\frac{0.4730 mol}{0.25 kg}=1.89 m$

C. 35.0 g of $C_2H_{6}O_2$ in 250.0 g of ethanol

Moles of $C_2H_{6}O_2$ = $\frac{35.0 g}{62g/mol}=0.5645 mol$

The mass of ethanol solvent is 250.0 g = 0.25 kg (1 g = 0.001 kg)

$m''=\frac{0.5645 mol}{0.25 kg}=2.26 m$

$m>m'''>m''$

The solution containing 35.0 g of $C_3H_8O$ added to 250.0 g of ethanol will yield the lowest freezing point

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A 200.0mL closed flask contains 2.000mol of carbon monoxide gas and 2.000mol of oxygen gas at the temperature of 300.0K. How man

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Answer:

0.400 moles of Oxygen

Explanation:

By employing the equation PV = nRT, the initial pressure of the flask can be calculated prior to the reaction, which leads to:

P = nRT/V

Where:

n signifies moles (4,000 moles: 2,000 moles of CO and 2,000 moles of H₂O)

R represents the gas constant (0.082 atm·L/mol·K)

T is the temperature (300.0 K)

V denotes volume (0.2000 L)

Substituting values results in P = 492.0 atm

To achieve a pressure reduction of 10.00%, the resulting pressure should be:

492.0 atm - 49.2 atm = 442.8 atm

Calculating with the new pressure under the same conditions gives the moles as:

n = PV/RT

n = 3,600 total moles

In the reaction:

2CO(g) + O₂(g) ⟶ 2CO₂(g)

The resulting moles are:

CO: 2,000 moles - 2X

O₂: 2,000 moles - X

CO₂: 2X

Where X accounts for the moles that react

Consequently, the total moles are:

4,000 moles - X = 3,600 moles

X = 0.400 moles

This indicates that the amount of oxygen needed for the reaction is 0.400 moles of Oxygen

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Clarification:

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In some ways, metallic bonding can be considered a variant of covalent bonding, but it is more communal—delocalized across numerous atoms—and electron deficient (there are more energy states than available electrons, which contributes to conductive traits). This implies that the term “metallic bond” might appear contradictory, akin to referring to a forest with a single tree.

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Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produ

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Answer: The molecular formula for the specified organic compound is $C_{18}H_{20}O_2$

Explanation:

The combustion reaction of a hydrocarbon comprising carbon, hydrogen, and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

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

The information provided includes:

Mass of $CO_2=39.61g$

Mass of $H_2O=9.01g$

From our knowledge:

Molar mass of carbon dioxide is 44 g/mol

Molar mass of water is 18 g/mol

For determining the amount of carbon:

In carbon dioxide weighing 44 g, 12 g of carbon is found.

Hence, in 39.61 g of carbon dioxide, $\frac{12}{44}\times 39.61=10.80g$ grams of carbon will be found.

For finding the mass of hydrogen:

In water weighing 18 g, 2 g of hydrogen can be found.

Thus, in 9.01 g of water, $\frac{2}{18}\times 9.01=1.00g$ grams of hydrogen will be present.

Mass of oxygen in the compound is given by (13.42) - (10.80 + 1.00) = 1.62 g

To derive the empirical formula, the following steps must be followed:

Step 1: Convert the indicated masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{10.80g}{12g/mole}=0.9moles$

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

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.62g}{16g/mole}=0.10moles$

Step 2: Calculating the ratio of moles of the respective elements.

To find the mole ratio, each mole value is divided by the smallest amount of moles calculated, which is 0.10 moles.

For Carbon = $\frac{0.9}{0.10}=9$

For Hydrogen = $\frac{1}{0.10}=10$

For Oxygen = $\frac{0.10}{0.10}=1$

Step 3:Using the mole ratios as subscripts.

The ratio of C: H: O = 9: 10: 1

Therefore, the empirical formula for the mentioned compound is $C_9H_{10}O$

To ascertain the molecular formula, it is necessary to find the valency, which is multiplied by each element to derive the molecular formula.

The equation to determine the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

Given the data:

Molecular formula mass = 268.34 g/mol

Empirical formula mass = 134 g/mol

Substituting the values into the aforementioned equation yields:

$n=\frac{268.34g/mol}{134g/mol}=2$

By multiplying this valency with each element's subscripts from the empirical formula, the results are:

$C_{(9\times 2)}H_{(10\times 2)}O_{(1\times 2)}=C_{18}H_{20}O_2$

Consequently, the molecular formula for the given organic compound is $C_{18}H_{20}O_2$ .

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