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3 years ago

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A 0.800 g sample of cuso4 (s was placed in calorimeter with 43.0g of water at an initial temp of 18.3 degrees celcious. as the c

uso(s dissolves in water, the temp increases to 20.2 degrees celcious. what value of /\h can be determined.

1 answer:

Serhud [2]3 years ago

6 0

Hello it's 92479x027430=3057292

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Find the boiling point?<br> 100. g of C2H6O2 dissolved in 200 g of H2O?

aleksklad [387]

Answer:

The correct answer is 104.13ºC

Explanation:

When a solute is added to a solvent, the boiling point of the solvent (Tb) increases. That is a colligative property. The increment in Tb (ΔTb) is given by the following expression:

ΔTb = Tb - Tbº= Kb x m

Where Tb and Tbº are the boiling points of the solvent in solution and pure, respectively; Kb is a constant and m is the molality of the solution.

In this problem, the solvent is water (H₂O). It is well known that water has a boiling point of 100ºC (Tb). The value of Kb for water is 0.512ºC/m. So, we have to calculate the molality of the solution (m):

m = moles of solute/Kg solvent

The solute is C₂H₆O₂ and we have to calculate the number of moles of this component by dividing the mass into the molecular weight (Mw):

Mw(C₂H₆O₂)= (2 x 12 g/mol) + (6 x 1 g/mol) + (2 x 16 g/mol)= 62 g/mol

⇒ moles of C₂H₆O₂ = mass/Mw = 100 g/(62 g/mol) = 1.613 moles

Now, we need the mass of solvent (H₂O) in kilograms, so we divide the grams into 1000:

200 g x 1 kg/1000 g = 0.2 kg

Finally, we calculate the molality as follows:

m = 1.613 moles of C₂H₆O₂/0.2 kg = 8.06 moles/kg = 8.06 m

The increment in the boiling point will be:

ΔTb = Kb x m = 0.512ºC/m x 8.06 m = 4.13ºC

So, the boiling point of pure water (Tbº=100ºC) will increase in 4.13ºC:

Tb= 100ºC+4.12ºC= 104.13ºC

5 0

3 years ago

A gas made up of atoms escapes through a pinhole times as fast as gas. Write the chemical formula of the gas.

Artemon [7]

A gas made up of atoms escapes through a pinhole 0.225times as fast as gas. Write the chemical formula of the gas.

Answer:

Explanation:

To solve this problem, we must apply Graham's law of diffusion. This law states that "the rate of diffusion or effusion of a gas is inversely proportional to the square root of its molecular mass at constant temperature and pressure".

Mathematically;

$\frac{r_{1} }{r_{2} } = \frac{\sqrt{m_{2} } }{\sqrt{m_{1} } }$

r₁ is the rate of diffusion of gas 1

r₂ is the rate of diffusion of gas 2

m₁ is the molar mass of gas 1

m₂ is the molar mass of gas 2

let gas 2 be the given H₂;

molar mass of H₂ = 2 x 1 = 2gmol⁻¹

rate of diffusion is 0.225;

i .e r1/r2 = 0.225

0.225 = √2 / √ m₁

0.225 = 1.414 / √ m₁

√ m₁ = 6.3

m₁ = 6.3² = 39.5g/mol

The gas is likely Argon since argon has similar molecular mass

5 0

3 years ago

Each of the following values was read on an instrument of measuring device. In each case the last digit was estimated. Tell what

Drupady [299]

Answer:

<h3>160 cm</h3>

Explanation:

6 0

3 years ago

A researcher wants to test the solubility (property of being dissolved) of salt in water as the temperature of the water increas

Leya [2.2K]

The researcher may first weight the beaker with water and then start to heat the water to a constant temperature, for example 30 °C and then start adding salt and stirring. He should add salt slowly until solid salt starts to become visible and the solution starts becoming cloudy. When this happens, he should quickly weigh the beaker. The increase in mass is the mass of salt dissolved at that temperature.
The procedure is then repeated but at an increased temperature until 5-6 temperatures have been tested.

3 0

3 years ago

Calculate the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water. The molal-freezing-point-depres

yulyashka [42]

<u>Answer:</u> The freezing point of solution is -0.454°C

<u>Explanation:</u>

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 2

$K_f$ = molal freezing point elevation constant = 1.86°C/m

$m_{solute}$ = Given mass of solute (KCl) = 5.0 g

$M_{solute}$ = Molar mass of solute (KCl) = 74.55 g/mol

$W_{solvent}$ = Mass of solvent (water) = 550.0 g

Putting values in above equation, we get:

$0-\text{Freezing point of solution}=2\times 1.86^oC/m\times \frac{5\times 1000}{74.55g/mol\times 550}\\\\\text{Freezing point of solution}=-0.454^oC$

Hence, the freezing point of solution is -0.454°C

3 0

3 years ago