Butter and grease are nonpolar compounds. what happens to greasy plates when dunked in plain water? what do you need to thoroughly clean a greasy dish and why?

- Since both butter and fat are not polar compounds, plain water (polar)  will be unable to clean all the fat from greasy dishes.

- What is needed to thoroughly clean a greasy dish is a detergent, since it makes it easier for the fat molecules to dissolve in the water.

Explanation:

Water is a molecule considered a universal solvent, given the physical-chemical characteristics that allow it to bind to a large number of molecules and chemical elements as it is a polar molecule. Lipids or fats, on the other hand, are non-polar molecules that cannot be diluted in water, but can be diluted in solvents such as alcohols.

The water (H₂O) is unable to drag the fat molecules on the surface of the plate, which is due to the difference in polarity of the water and fat molecules.

Lipids are non-polar and immiscible molecules in H₂O because they are hydrocarbon chains with single or double bonds that do not show polarity, and their adhesion to the H₂O molecule is not possible.

To remove grease from a greasy dish, a detergent should be applied.

Detergents and soaps are chemical substances that have the property of emulsifying lipids and fats, because they have a hydrophilic portion -with an affinity for water- and a lipophilic one.

When a fat molecule is chemically emulsified it becomes soluble in water -the hydrophilic portion acts with H₂O- which makes it easier for a detergent to facilitate the cleaning of greasy dishes.

c)red bank

explanation:

red bank is a borough in monmouth county, new jersey, incorporated in 1908 and located on the navesink river, the area's original transportation route to the ocean and other ports.

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol