Answer:

The formula for osmotic pressure is:

where is osmotic pressure, is van't Hoff's factor, molarity, is Ideal gas constant, and T is Temperature.

= 132 atm

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

Substituting the values in the above equation we get,

So, the molarity of the solution is .

A solution containing 292 g of Mg(NO3)2 per liter has a density of 1.108 g/mL. The molality of the solution is:A) 2.00 mB) 1.77 mC) 6.39 mD) 2.41 mE) none of these

Explain the arrangement of the first 20 elements in the periodic table. Please help! Will give brainliest!

4.What volume of 0.120 M HNO3(aq) is needed tocompletely neutralize 150.0 milliliters of 0.100 MNaOH(aq)?A. 62.5 mLB. 125 mlC.180. mLD. 360. mL

Calculate the mass percent of oxygen in KMnO4.

What is the molarity of the potassium hydroxide if 27.20 mL of KOH is required to neutralize 0.604 g of oxalic acid, H2C2O4?H2C2O4(aq)+2KOH(aq)→K2C2O4(aq)+2H2O(l)

Explain the arrangement of the first 20 elements in the periodic table. Please help! Will give brainliest!

4.What volume of 0.120 M HNO3(aq) is needed tocompletely neutralize 150.0 milliliters of 0.100 MNaOH(aq)?A. 62.5 mLB. 125 mlC.180. mLD. 360. mL

Calculate the mass percent of oxygen in KMnO4.

What is the molarity of the potassium hydroxide if 27.20 mL of KOH is required to neutralize 0.604 g of oxalic acid, H2C2O4?H2C2O4(aq)+2KOH(aq)→K2C2O4(aq)+2H2O(l)

How many atoms of each element are in

the equation?

**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

**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

**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:**

where,

= Henry's constant =

= partial pressure of hydrogen sulfide gas = 2.42 atm

**Putting values in above equation, we get:**

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

**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:**

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

**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.

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!!!

**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