Paper chromatography is a technique used to separate molecules based upon their size and solubility in a particular solvent. If pigments from a particular species of plant are extracted and subjected to paper chromatography, which of the following results is most likely?
Paper chromatography would separate the pigments into several bands that appear green or yellow/orange.

Paper chromatography would separate the pigments into several bands that appear green or yellow/orange.

Explanation:

Chloroplasts have a mixture of pigments with different colors: intense green chlorophyll-a, green chlorophyll-b, yellow carotenes and orange-yellow xanthophylls, in different proportions. All these substances have a different degree of solubility in non-polar solvents, which allows their separation when a solution of them ascends by capillarity through a porous paper strip, vertically arranged on a film of an organic solvent, where the most soluble move faster, and the less soluble move less on the filter paper strip. Therefore, several bands of different colors will appear that will be more or less distant from the solution according to the greater or lesser solubility of the pigments.

dishwashers don't have rights

explanation:

dishwashers are meant for one thing and one thing only: dishwashing. they belong in the kitchen making sandwiches and cleaning up for the man of their household. if your dishwasher starts acting up just slap it and wait a minute.

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