A cypress tree is the sprout stage. what would you expect to notice when observing this plant?

Chemistry

1 answer:

Anna007 [38]3 years ago

8 0

Answer:

A mature stem that has thick bark

Explanation:

i dont know but that is what i put on my test but if you had your awnser right tell me

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chemistry and biology are so closely related that sometimes the fields overlap. which of the following experiments might both ch

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The answer to your question is letter A. study how cholesterol is broken down in the body. Biology is a study of life which deals with cells, organelles, and other essential parts of the body or organism. Chemistry is a study about chemicals and how it reacts to natural state. In studying about cholesterol in the body, the presence of biologist and chemists will study how the body reacts with the chemicals and substances that were stored in the cholesterol.

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3 years ago

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Calculate the molality of a solution containing 22.75 g of glycerol (C3H8O3) in 79.6 g of ethanol (C2H5OH).

3241004551 [841]

Answer:

3.11 mol/kg

Explanation:

Molality M = number of moles of solute, n/mass of solvent, m

To calculate the number of moles of glycerol (C₃H₈O₃) in 22.75 g of glycerol, we find its molar (molecular) mass, M',

So, M' = 3 × atomic mass of carbon + 8 × atomic mass hydrogen + 3 × atomic mass of oxygen

= 3 × 12 g/mol + 8 × 1 g/mol + 3 × 16 g/mol = 36 g/mol + 8 g/mol + 48 g/mol = 92 g/mol.

So, number of moles of glycerol, n = m'/M' where m' = mass of glycerol = 22.75 g and M' = molecular mass of glycerol = 92 g/mol

So, n = m'/M'

n = 22.75 g/92 g/mol

n = 0.247 mol

So, the molality of the solution M = n/m

Since m = mass of ethanol = 79.6 g = 0.0796 kg, substituting the value of n into the equation, we have

M = 0.247 mol/0.0796 kg

M = 3.11 mol/kg

So, the molality of the solution is 3.11 mol/kg.

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3 years ago

Draw the organic product formed when the structure shown below undergoes a reaction with nanh2. there is no aqueous workup in th

blagie [28]

Answer:

Prop-1-yn-1-ide

Explanation:

In this case, we will have an acidic hydrogen upon the left carbon in the alkyne. In the presence of a base, we can remove this hydrogen. In this reaction the base will be $NH_2^-$ . The amidure ion will remove the hydrogen of the alkyne to produce a carboanion. (See figure)