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2 years ago

A 250 ml flask contains 3.4 g of neon gas at 45°c. Calculate the pressure of the neon gas inside the flask.

Chemistry

1 answer:

Rudiy272 years ago

7 0

Answer:

The answer to your question is P = 17.73 atm

Explanation:

V = 250 ml = 0.25 liters (L)

mass = 3.4 g

Temperature = 45°C = 45 + 273 = 318°K

Pressure = ?

R = 0.082 atm L / mol°K

Atomic mass Ne = 20.18 g

Process

1.- Calcule moles of Neon

20.18 g of Ne ------------------ 1 mol

3.4 g of Ne ------------------- x

x = (3.4 x 1)/20.18

x = 0.17 moles of Neon

2.- Find pressure

Ideal gases formula PV = nRT

$P = \frac{nRT}{V}$

$P = \frac{(0.17)(0.082)(318)}{0.25}$

$P = \frac{4.433}{0.25}$

P = 17.73 atm

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A 0.286-g sample of gas occupies 125 ml at 60. cm of hg and 25°

irga5000 [103]

Using the Equation: PV=nRT
Where P is the pressure 60 cmHg or 600 mmHg or 600/760= 0.789 atm
V is the volume 125 ml or 0.125 L, n is the number of moles, R is a constant 0.082057, and T is temperature 25 °C or 298 K;
Therefore:
0.789 × 0.125 = n × 0.082057 × 298
n = 0.0987/24.45
= 0.004036 mol
0.004036 mole has a mass of 0.286 g
Hence; 1 mole has a mass of 0.286/0.004036
= 70.8 g /mol
Therefore the molar mass of the gas is 71 g/mol (2 sfg)

4 0

2 years ago

When of benzamide are dissolved in of a certain mystery liquid , the freezing point of the solution is less than the freezing po

SOVA2 [1]

The given question is incomplete. The complete question is as follows.

When 70.4 g of benzamide ( $C_{7}H_{7}NO$ ) are dissolved in 850 g of a certain mystery liquid X, the freezing point of the solution is $2.7^{o}C$ lower than the freezing point of pure X. On the other hand, when 70.4 g of ammonium chloride ( $NH_{4}Cl$ ) are dissolved in the same mass of X, the freezing point of the solution is $9.9^{o}C$ lower than the freezing point of pure X.

Calculate the Van't Hoff factor for ammonium chloride in X.

Explanation:

First, we will calculate the moles of benzamide as follows.

Moles of benzamide = $\frac{mass}{\text{Molar mass of benzamide}}$

= $\frac{70.4 g}{121.14 g/mol}$

= 0.58 mol

Now, we will calculate the molality as follows.

Molality = $\frac{\text{moles of solute (benzamide)}}{\text{solvent mass in kg}}$

= $\frac{0.58 mol}{0.85 kg}$

= 0.6837

It is known that relation between change in temperature, Van't Hoff factor and molality is as follows.

dT = $i \times K_{f} \times m$ ,

where, dT = change in freezing point = $2.7^{o}C$

i = van't Hoff factor = 1 for non dissociable solutes

$K_{f}$ = freezing point constant of solvent

m = 0.6837

Therefore, putting the given values into the above formula as follows.

dT = $i \times K_{f} \times m$ ,

$2.7^{o}C = 1 \times K_{f} \times 0.6837 m$

$K_{f}$ = 3.949 C/m

Now, we use this $K_{f}$ value for calculating i for $NH_{4}Cl$

So, moles of ammonium chloride are calculated as follows.

Moles of $NH_{4}Cl = \frac{70.4 g}{53.491 g/mol}$

= 1.316 mol

Hence, calculate the molality as follows.

Molality = $\frac{1.316 mol}{0.85 kg}$

= 1.5484

It is given that value of change in temperature (dT) = $9.9^{o}C$ . Thus, calculate the value of Van't Hoff factor as follows.

dT = $i \times K_{f} \times m$

$9.9^{o}C = i \times 3.949 C/m \times 1.5484 m$

i = 1.62

Thus, we can conclude that the value of van't Hoff factor for ammonium chloride is 1.62.

5 0

2 years ago

The dipole moment (μ) of HBr (a polar covalent molecule) is 0.838D (debye), and its percent ionic character is 12.4 % . Estimate

myrzilka [38]

When two electrical charges, of opposite sign and equal magnitude, are separated by a distance, a dipole is established. The size of a dipole is measured by its dipole moment (μμ). Dipole moment is measured in Debye units, which is equal to the distance between the charges multiplied by the charge (1 Debye equals 3.34×10−30Cm3.34×10−30Cm). The dipole moment of a molecule can be calculated by Equation 1.11.1:

μ = qr

where

μ⃗ μ→ is the dipole moment vector qiqi is the magnitude of the ithith charge, and r⃗ ir→i is the vector representing the position of ithith charge.

r = μ/q

r = [0.838D(3.34×10−30 C⋅m/ 1D)]/ (1.6×10−19 C) *0.124
 r = 1.41 x10^-10 m

7 0

2 years ago

Read 2 more answers

3) Calculate the percent by mass of 3.55 g NaCl dissolved in 88 g water.

kotykmax [81]

Answer:

The percent by mass of 3.55 g NaCl dissolved in 88 g water is 3.88%

Explanation:

When a solute dissolves in a solvent, the mass of the resulting solution is a sum of the mass of the solute and the solvent.

A percentage is a way of expressing a quantity as a fraction of 100. In this case, the percentage by mass of a solution is the number of grams of solute per 100 grams of solution and can be represented mathematically as:

$Percent by mass=\frac{mass of solute}{mass of solution} *100$

In this way it allows to precisely establish the concentration of solutions and express them in terms of percentages.

In this case:

mass of solute: 3.55 g
mass of solution: 3.55 g + 88 g= 91.55 g

Replacing:

$Percentbymass=\frac{3.55}{91.55}*100$

Percent by mass= 3.88%

The percent by mass of 3.55 g NaCl dissolved in 88 g water is 3.88%

5 0

2 years ago

Which two scenarios illustrate the relationship between pressure and volume as described by Boyle’s law?

Kruka [31]

Answer:

2) The volume of an underwater bubble increases as it rises and the pressure decreases

hope it was useful for you

stay at home stay safe

5 0

2 years ago