Which statement accurately describes the rock layers

a solution is made by completely dissolving 90. grams of kno3 in 100 grams of water in a beaker. the temperature of this solutio

jolli1 [7]

Pressure has no effect on the solubility of KNO3 in water. This is because it is solid in liquid type of solution. In solid in liquid type of solution, solid is solute (minor component), liquid is solvent (major component). For solid in liquid type of solutions, solubility is independent of pressure.

On other hand, pressure has a pronounced effect on the solubility of gas in liquid type solutions. In such system, gas is solute (minor component) and liquid is solvent (major component). Example of such solution is aerated water. Herein, CO2 is dissolved in water. In such gas in liquid type of solutions, solubility increases with increasing pressure.

3 0

3 years ago

A sample of table sugar (sucrose, C12H22O11) has a mass of 1.202 g.

Archy [21]

Answer:

a) 0.003512 moles

b) Moles C= 0.04214 moles carbon

Moles H = 0.07726 moles hydrogen

Moles O = 0.03863 moles of oxygen

c) C atoms = 2.54 *10^22 carbon atoms

H atoms = 4.65 *10^22 hydrogen atoms

O atoms = 2.33 *10^22 oxygen atom

Explanation:

Step 1: Data given

Mass of sucrose = 1.202 grams

Molar mass of sucrose = 342.3 g/mol

Step 2: Calculate moles of sucrose

Moles sucrose = Mass sucrose / molar mass sucrose

Moles sucrose = 1.202 grams / 342.3 g/mol

Moles sucrose = 0.003512 moles

Step 3: Calculate moles of each element

For 1 mol of C12H22O11 we have 12 moles of carbon, 22 moles of hydrogen and 11 moles of oxygen

Moles C: 12*0.003512 = 0.04214 moles carbon

Moles H: 22* 0.003512 = 0.07726 moles hydrogen

Moles O: 11* 0.003512 = 0.03863 moles of oxygen

Step 4: Calculate the number of atoms

C atoms = 6.022 *10^23 / mol * 0.04214 moles = 2.54 *10^22 atoms carbon

H atoms = 6.022 * 10^23 / mol * 0.07726 moles = 4.65 *10^22 atoms H

O atoms = 6.022 * 10^23 / mol * 0.03863 moles = 2.33 *10^22 atoms O

0.003512 moles of sucrose containse 6.022 *10^23 * 0.003512 = 2.11 * 10^21 sucrose molecules

7 0

3 years ago

pressure has little effect on the solubility of liquids and solids because they are almost incompressible. True or False

slavikrds [6]

Pressure has little effect on the solubility of liquids and solids because they are almost incompressible True.

Liquids and solids show little change in solubility with changes in pressure. As expected, gases increase in solubility with increasing pressure. Henry's Law states that the solubility of a gas in a liquid is directly proportional to the pressure of that gas above the surface of the solution.

External pressure has little effect on liquid and solid solubility. In contrast, the solubility of a gas increases as the partial pressure of the gas above the solution increases.

Solubility is a measure of the concentration of dissolved gas particles in a liquid and is a function of gas pressure. Increasing the gas pressure increases the number of collisions and increases the solubility, and decreasing the pressure decreases the solubility.

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5 0

1 year ago

Give the formula for the alkane containing 500 carbons.

Fiesta28 [93]

Answer:

1002

Explanation:

$C_{2n}$ $H_{2n+2}$

500 multiply by 2 add 2

5 0

3 years ago

D( )-Camphor is treated with 5-methoxytryptamine in the presence of sodium cyanoborohydride to generate a new molecule, as a sin

Svetach [21]

The reducing agent can approach the carbonyl face of camphor by forming a one carbon bridge (known as an exo attack) or a two carbon bridge (termed endo).

The two resultant stereoisomers are known as isoborneol and borneol (from exo attack) (from endo attack). Gas chromatography (GC) analysis may be used to calculate the ratio of each isomeric alcohol in the mixture. Unfortunately, IR analysis does not permit this.

The stereochemistry of the reaction is regulated in stiff cyclic compounds like camphor and norcamphor by protecting one side of the carbonyl group from the reagent's assault. The hydrogen atom is added to the endo side, creating the exo alcohol isoborneol, while the methyl groups on the one-carbon bridge of camphor screen the approach of the hydride from the "top" or exo side of the two-carbon bridge. You will be asked to guess the main isomeric alcohol created by the norcamphor hydride reduction later in the lab report.

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4 0

2 years ago