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ddd [48]
3 years ago
15

What is the volume of an object whose dimensions are 5.54 cm, 10.6 cm, and 199 cm?

Chemistry
1 answer:
cluponka [151]3 years ago
3 0
Volume:

a x b  x c 

Therefore:

5.54 x 10.6 cm x 199 => 11,686.076 cm³
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Classify each property as intensive or extensive. Color, temperature, volume, hardness, mass, boiling point, length, density
mr_godi [17]

Answer:

Intensive properties can be used to help identify a sample because these characteristics do not depend on the amount of sample, nor do they change according to conditions.

Explanation:

Intensive properties are bulk properties, which means they do not depend on the amount of matter that is present. Examples of intensive properties include:

Boiling Point

Density

State of Matter

Color

Melting Point

Odor

Temperature

Refractive Index

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3 0
3 years ago
Read 2 more answers
Can you draw the peptide structure at pH 7.0 (including peptide bonds)
ELEN [110]

Answer:

See image below

Explanation:

The image is labeled according to the sequence N'-Trp-Ser-Asg-Gly-Cys-His-COOH' which means that in the main chain of the peptide, the amino group of the Tryptophan and the carboxylic group of the Histidine are free and thus its charge depends on the pH; other groups that rely on the pH are the side groups of the Cysteine and the Histidine.

Overall, ionizable groups in this peptide are:

  • Amino Group of the Tryptophan (pKa = 9.39)
  • SH group of the Cysteine (pKa = 8.18)
  • Secondary amine of the Histidine (pKa = 6.00)
  • Carboxylic Group of the Histidine (pKa = 1.82)

Then, the amino group of Trp and SH group of Cys are protonated since the peptide is at a pH below the pKa. The secondary amine of the Histidine is deprotonated because the pH is greater than the pKa, as well with its carboxylic acid group.

7 0
3 years ago
A compound has a chemical composition of 97.6 g carbon, 4.9 g hydrogen, 52 g oxygen, 45.5 g nitrogen. What is the empirical form
Alekssandra [29.7K]
First  step  is  to  calculate  the  mole  of each   element
that  is;
carbon  97.6/12=8.13moles
hydrogen= 4.9/1=4.9  moles
oxygen 52/16=3.25  moles
nitrogen=45.5/14=3.25 moles

step  two  is  to  calculate  the mole ratio  by   dividing with  the  smallest  number of  moles
that  is  divide  each  mole  with  3.25moles
 carbon=8.13/3.25 =5/2
hydrogen=4.9/3.25= 3/2
oxygen=3.25/3.25=1
nitrogen=3.25/3.25=1
step  3;  multiply  all  the  mole  ratio by  2  to  remove  the  fraction
carbon=5/2  x2 =5
hydrogen=3/2 x2=3
oxygen=1 x2=2
nitrogen  =1x2=2
therefore  the  empirical  formula  is C5H3O2N2(answer c)
8 0
3 years ago
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Ymorist [56]

Answer:

I and III

Explanation:

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Answer:

Group 17, 2, 7, 7 electrons

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