Answer:
The sugar in DNA is deoxyribose. ... Nucleotides in DNA contain four different nitrogenous bases: Thymine, Cytosine, Adenine, or Guanine. There are two groups of bases: Pyrimidines: Cytosine and Thymine each have a single six-member ring.
Instability of an atoms nucleus can result from an excess of either neutrons or protons . So neutrons and protons .
Answer:
1335.12 mL of H2O
Explanation:
To calculate the mililiters of water that the solution needs, it is necessary to know that the volume of the solution is equal to the volume of the solute (NaOH) plus the volume of the solvent (H2O).
From the molarity formula we can first calculate the volume of the solution:
The volume of the solution as we said previously is:
Solution volume = solute volume + solvent volume
To determine the volume of the solute we first obtain the grams of NaOH through the molecular weight formula:
Now with the density of NaOH the milliliters of solute can be determined:
Having the volume of the solution and the volume of the solute, the volume of the solvent H2O can be calculated:
Solvent volume = solution volume - solute volume
Solvent volume = 1429 mL - 93.88 mL = 1335.12 mL of H2O
Answer:
ane, al, keto
ol, al, keto
ol, al, one
ol, ane, one.
Explanation:
The suffix –ol is used in organic chemistry principally to form names of organic compounds containing the hydroxyl (–OH) group, mainly alcohols (also phenol). The suffix was extracted from the word alcohol. The suffix also appears in some trivial names with reference to oils (from Latin oleum, oil).
Functional group is a ketone, therefore suffix = -one
Hydrocarbon structure is an alkane therefore -ane
The longest continuous chain is C5 therefore root = pent
The first point of difference rule requires numbering from the left as drawn to make the ketone group locant 2-
pentan-2-one or 2-pentanone
CH3CH2CH2C(=O)CH3
Take a hypothetical sample of exactly 100 grams of the solution.
(16g urea) / (60.06 g urea/mol) = 0.2664 mol urea
((100 g total) - (16g urea)) = 84.0 g H2O = 0.0840 kg H2O
(0.2664 mol) /0.0840 (kg) = 3.17143mol/kg = 3.18m urea
