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Hoochie [10]
3 years ago
8

A sample of barium nitrate is placed into a jar containing water. The mass of the barium nitrate sample is 27g. Assume the water

is at 20 degrees C and the resulting barium nitrate solution is saturated. What mass of water is present in the jar?
Chemistry
2 answers:
lora16 [44]3 years ago
8 0

<u>Answer:</u> The mass of water present in the jar is 297.7 grams.

<u>Explanation:</u>

We are given:

Mass of barium nitrate sample = 27 g

Solubility is defined as the maximum amount of solute that can be dissolved in a solvent at equilibrium.

The solubility of barium nitrate at 20°C is 9.02 g per 100 g of water

To calculate the mass of solution in which given amount of barium nitrate is dissolved, we apply unitary method:

9.02 g of barium nitrate is dissolved in 100 g of water

So, 27 g of barium nitrate will be dissolved in \frac{100}{9.07}\times 27=297.7g of water

Hence, the mass of water present in the jar is 297.7 grams.

34kurt3 years ago
5 0
So we have Barium nitrate with a solubility of 8.7g in 100g water at 20°C.

using that relation
i.e.
8.7g (barium nitrate) =100g (water)
1g barium nitrate = 100/8.7 g water

27g barium nitrate = (100/ 8.7 ) × 27
= 310.34 g

therefore,
you need 310.34g of water is in the jar.

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A 0.532 mol sample of SO2 gas requires 52.3 s to effuse through a tiny hole. Under the same conditions, how long will it take 0.
scoundrel [369]

Answer:

41.3 s

Explanation:

Let t₁ represent the time taken for SO₂ to effuse.

Let t₂ represent the time taken for Ar to effuse.

Let M₁ represent the molar mass of SO₂

Let M₂ represent the molar mass of Ar

From the question given above,

Time taken (t₁) for SO₂ = 52.3 s

Time taken (t₂) for Ar =?

Molar mass (M₁) of SO₂ = 32 + (16×2) = 32 + 32 = 64 g/mol

Molar mass (M₂) of Ar = 40 g/mol

Finally, we shall determine the time taken for Ar to effuse by using the Graham's law equation as shown below:

t₂ / t₁ = √(M₂ / M₁)

t₂ / 52.3 = √(40 / 64)

t₂ / 52.3 = √0.625

t₂ / 52.3 = 0.79

Cross multiply

t₂ = 52.3 × 0.79

t₂ = 41.3 s

Thus, the time taken for the amount of Ar to effuse is 41.3 s

3 0
3 years ago
The density of water is 1.00g/ml at 4 degrees Celsius.How many water molecules are present in 2.56ml of water at this temperatur
vivado [14]

Answer:

8.55x10^22 molecules

Explanation:

From the question given, the following data were obtained:

Density = 1g/mL

Volume = 2.56mL

Mass =?

Density = Mass /volume

Mass = Density x volume

Mass = 1 x 2.56

Mass = 2.56g

Now let us convert this mass (i.e 2.56g) of water to mole

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Mass of H2O = 2.56g

Number of mole of H2O=? Number of mole = Mass /Molar Mass

Number of mole of H2O = 2.56/18

Number of mole of H2O = 0.142mol

From Avogadro's hypothesis, 1mole of any substance contains 6.02x10^23 molecules. This means that 1mole of H2O contains 6.02x10^23 molecules.

Now if 1mole of H2O contains 6.02x10^23 molecules, then 0.142mol of H2O will contain = 0.142 x 6.02x10^23 = 8.55x10^22 molecules

6 0
3 years ago
The vapor pressure of substance X is 100. mm Hg at 1080.°C. The vapor pressure of substance X increases to 600. mm Hg at 1220.°C
artcher [175]

Explanation:

The given data is as follows.

         P_{1} = 100 mm Hg or \frac{100}{760}atm = 0.13157 atm

         T_{1} = 1080 ^{o}C = (1080 + 273) K = 1357 K

         T_{2} = 1220 ^{o}C = (1220 + 273) K = 1493 K

         P_{2} = 600 mm Hg or \frac{600}{760}atm = 0.7895 atm

          R = 8.314 J/K mol

According to Clasius-Clapeyron equation,

                   log(\frac{P_{2}}{P_{1}}) = \frac{\Delta H_{vap}}{2.303R}[\frac{1}{T_{1}} - \frac{1}{T_{2}}

            log(\frac{0.7895}{0.13157}) = \frac{\Delta H_{vap}}{2.303 \times 8.314 J/mol K}[\frac{1}{1357 K} - \frac{1}{1493 K}]

          log (6) = \frac{\Delta H_{vap}}{19.147}[\frac{(1493 - 1357) K}{1493 K \times 1357 K}]

                0.77815 = \frac{\Delta H_{vap}}{19.147J/K mol} \times 6.713 \times 10^{-5} K

              \Delta H_{vap} = 2.219 \times 10^{5} J/mol

                                   = 2.219 \times 10^{5}J/mol \times 10^{-3}\frac{kJ}{1 J}

                                    = 221.9 kJ/mol

Thus, we can conclude that molar heat of vaporization of substance X is 221.9 kJ/mol.

4 0
3 years ago
State the five the five basic assumptions of the kinetic-molecular theory.
Ivan

Answer:

The primary assumptions are as follows:

Any gas is a collection of innumerable number of minuscule particles which are known as molecules according to Avogadro’s law.

There are no forces of attraction or repulsion among the particles or between the molecules and the surroundings.

The gas particles are always at straight, rapid, fast & random motion resulting in inevitable collisions with other particles and the surroundings that changes direction of motion.

Since the particle are spherical, solid and elastic the collisions involving them are elastic in nature as well i.e their kinetic energy is conserved even after collisions.

The total kinetic energy of the particles is proportional to the absolute temperature.

In some books two other assumptions are given as well:

1. The size or area of each particle is negligible compared to that of the container.

2. Pressure of gas is result of the continuous clash of the particles with the wall of the container.

or

The simplest kinetic model is based on the assumptions that: (1) the gas is composed of a large number of identical molecules moving in random directions, separated by distances that are large compared with their size; (2) the molecules undergo perfectly elastic collisions (no energy loss) with each other and with the walls of the container, but otherwise do not interact; and (3) the transfer of kinetic energy between molecules is heat. These simplifying assumptions bring the characteristics of gases within the range of mathematical treatment.

Such a model describes a perfect gas and is a reasonable approximation to a real gas, particularly in the limit of extreme dilution and high temperature. Such a simplified description, however, is not sufficiently precise to account for the behaviour of gases at high densities.

Based on the kinetic theory, pressure on the container walls can be quantitatively attributed to random collisions of molecules the average energy of which depends upon the gas temperature. The gas pressure can therefore be related directly to temperature and density. Many other gross properties of the gas can be derived, such as viscosity, thermal and electrical conductivity, diffusion, heat capacity, and mobility. In order to explain observed deviations from perfect gas behaviour, such as condensation, the assumptions must be appropriately modified. In doing so, considerable insight has been gained as to the nature of molecular dynamics and interactions.

7 0
2 years ago
We discussed the different types of intermolecular forces in this lesson. Which type would you expect to find in CO2?
g100num [7]

Answer:

Dispersion forces.

Explanation:

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CO contains two C-O bonds. They cancel each other out because of the dipoles point in opposite directions. Although, CO2 contains polar bonds, it is known as a nonpolar molecule. So, the only intramolecular forces which CO2 having are London dispersion forces.

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