The osmotic pressure, π, of a solution of glucose is 132 atm . find the molarity of the solution at 298 k.

Answers

Answer 1
Answer:

The formula for osmotic pressure is:

\Pi = iMRT

where \Pi is osmotic pressure, i is van't Hoff's factor, M molarity, R is Ideal gas constant, and T is Temperature.

\Pi = 132 atm

The van't Hoff's factor for glucose, i = 1

R = 0.08206 Latmmol^(-1)K^(-1)

T = 298 K

Substituting the values in the above equation we get,

132 atm = 1* M* 0.08206 Latmmol^(-1)K^(-1)* 298

M = (132 atm)/(1* 0.08206 Latmmol^(-1)K^(-1)* 298) = 5.4797 molL^(-1) \simeq 5.48 molL^(-1)

So, the molarity of the solution is 5.48 molL^(-1).


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6 CO2 + 6 H2O
How many atoms of each element are in
the equation?

Answers

Answer:

There are 6 Carbon dioxides, and 6 waters, but there are 6 carbons, 18 oxygens, and 12 hydrogens.

Explanation:

Answer:

6 carbon atoms

18 oxygen atoms

12 hydrogen atoms

Calculate the molality of a 35.4 % (by mass) aqueous solution of phosphoric acid (H3PO4) (35.4 % means 35.4 g of H3PO4in 100 g of solution)

Answers

Answer:

3.6124 m/kg

Explanation:

Molality is calculated as moles of solute (mol) divided by kilogram of solvent (kg). Here, we can find these numbers by using the 35.4%, which gives us  35.4 g of H3PO4 and 100 g of solution to work with.

To go from grams to moles for the phosphoric acid, you need to find the molar mass of the compound or element and divide the grams of the compound or element by that molar mass.

Here, the molar mass for phosphoric acid is 97.9952 g/mol. The equation would look like this:

35.4 g x 1 mol / 97.9952 g = 0.3612422 mol

Next, the 100 g of solvent can easily be converted to 0.1 kg of solvent.

To find the molality, divide the moles of solute and kilograms of solution.

0.3612422 mol / 0.1 kg = 3.6124 m/kg

At 25.0 ⁰C the henry's law constant for hydrogen sulfide(H2S) gas in water is 0.087 M/atm. Caculate the mass in grams of H2S gas that can be dissolved in 400.0 ml of water at 25.00 C and a H2S partial pressure of 2.42atm.

Answers

Answer: The mass of hydrogen sulfide that can be dissolved is 2.86 grams.

Explanation:

Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the gas.

To calculate the molar solubility, we use the equation given by Henry's law, which is:

C_(H_2S)=K_H* p_(liquid)

where,

K_H = Henry's constant = 0.087M/atm

p_(H_2S) = partial pressure of hydrogen sulfide gas = 2.42 atm

Putting values in above equation, we get:

C_(H_2S)=0.087M/atm* 2.42atm\n\nC_(H_2S)=0.2105M

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

\text{Molarity of the solution}=\frac{\text{Mass of solute}* 1000}{\text{Molar mass of solute}* \text{Volume of solution (in mL)}}

We are given:

Molarity of solution = 0.2105 M

Molar mass of hydrogen sulfide = 34 g/mol

Volume of solution = 400.0 mL

Putting values in above equation, we get:

0.2105M=\frac{\text{Mass of hydrogen sulfide}* 1000}{34g/mol* 400.0mL}\n\n\text{Mass of }H_2S=(0.2105* 34* 400)/(1000)=2.86g

Hence, the mass of hydrogen sulfide that can be dissolved is 2.86 grams.

Structure and function of cellulose?

Answers

Answer:

Cellulose is a major component of tough cell walls that surround plant cells, and it's what makes plant stems, leaves, and branches so strong. ... This forms long, cable-like structures, which combine with other cellulose molecules and is what produces such a strong support structure.

Explanation:

Answer:

Cellulose, a tough, fibrous, and water-insoluble polysaccharide, plays an integral role in keeping the structure of plant cell walls stable.

Explanation:

Cellulose chains are arranged in microfibrils or bundles of polysaccharide that are arranged in fibrils (bundles of microfibrils), which in turn make up the plant cell wall.

A 0.7 ft diameter hole forms in a tank containing butane at 19 atmg and 76 degrees Fahrenheit. Determine the maximum possible mass flow rate through this leak in lb m / min, if the external pressure is 1 atm.

Answers

Answer:

Q = 3,534.4 lbm/s = 212,062 lbm/min

Explanation:

Mass flowrate of discharge or leakage mass flowrate (Q) is given as

Q = AC₀√(2ρgP)

A = Cross sectional Area of leakage = (πD²/4) = (π×0.7²)/4

A = 0.385 ft²

C₀ = discharge coefficient = 0.98 (For maximum discharge flow rate, the flow is turbulent with discharge coefficient within 1% of 0.98)

ρ = density of butane at 76°F = 35.771 lbm/ft³

g = acceleration due to gravity = 32.2 lbm.ft/lbf.s²

P = Gauge Pressure in the tank = (absolute pressure) - (external pressure) = 19 - 1 = 18 atm = 38091.9 lbf/ft²

Q = AC₀√(2ρgP)

Q = (0.385)(0.98)√(2×35.771×32.2×38091.9)

Q = 3,534.4 lbm/s = 212,062 lbm/min

Hope this Helps!!!

Write the balanced reaction and solubility product expression (KSP) for dissolving silver chromate: Ag2CrO4(s). Include all charges, stoichiometric coefficients, and phase subscripts.

Answers

Answer:

2Ag⁺ (aq)  + CrO₄⁻² (aq) ⇄  Ag₂CrO₄ (s) ↓

Ksp = [2s]²  . [s] → 4s³

Explanation:

Ag₂CrO₄ → 2Ag⁺  + CrO₄⁻²

Chromate silver is a ionic salt that can be dissociated. When we have a mixture of both ions, we can produce the salt which is a precipitated.

2Ag⁺ (aq)  + CrO₄⁻² (aq) ⇄  Ag₂CrO₄ (s) ↓ Ksp

That's the expression for the precipitation equilibrium.

To determine the solubility product expression, we work with the Ksp

Ag₂CrO₄ (s)  ⇄ 2Ag⁺ (aq)  + CrO₄⁻² (aq)   Ksp

                          2 s                 s

Look the stoichiometry is 1:2, between the salt and the silver.

Ksp = [2s]²  . [s] → 4s³