Draw the structure of the peptide DTLH, showing the backbone and side-chain atoms, at its isoelectric point. Draw the molecule on the canvas by choosing buttons from the Tools (for bonds and charges), Atoms, and Templates toolbars. Draw peptide molecule as zwitter ion.
Answer:
Explanation:
A peptide is any class of organic compounds composed of different numbers of amino acids in which the amine of one is reacted with the carboxylic acid of the next to form an amine bond. The peptide of DTLH is composed of the following amino acids:
Asp-Thr-Leu-His
Their structures are first drawn out in the image attached below. This is followed by the isoelectric structure of the peptide DTLH
The Water Cycle: Lesson
Mckenze Jelks
Which example best shows how the water cycle carries on energy transfer?
Choose the correct answer.
O a flooding river depositing silt on a floodplain
O a warm ocean current warming the air above it
O ocean water depositing sand particles on a shore
O water seeping through the soil and dissolving salts
what is the difference between fluorine and oxygen?
Answer:
Continuing on across the periodic table we see that fluorine is the next element after oxygen. ... Rather than forming seven bonds fluorine only forms a single bond for basically the same reasons that oxygen only forms two bonds. Hydrogen fluoride, HF, has one bond, but four centers of electron density around the fluorine.
nitrogen atom in NH3 has ___ shared electrons
Answer:
3
Explanation:
nitrogen atom in NH3 has 3 shared electrons.
How many moles are in 3.30 x 1024 molecules of N2I6
Divide the number of molecules N by avogadro's constant.
Let n be the number of moles then,
[tex]n=\frac{N}{N_A}=\frac{3.3\cdot10^{24}}{6.022\cdot10^{23}}=5.48\mathrm{mol}[/tex]
Hope this helps.
HURRY PLEASE
If 12 grams of sodium reacts with 16 grams of chlorine, how much sodium chloride is Formed?
A. 4 g
B. 12 g
C. 28 g
D. 30 g
Answer:
6.576 g NaCl
Explanation:
2Na + Cl2 --> 2 NaCl
Step 1: Find out moles of Na and Cl to determine the limiting reactant:
moles Na = 12/23 = 0.52 moles Na
moles Cl2 = 16/71=0.225 moles Cl2
ratio Na:Cl is 2.3:1 so Cl is the limiting reactant.
Step 2: How many moles of NaCl are formed:
moles NaCl = moles Cl2 x (1 mol NaCl/2 moles Cl2) = 0.225/2 = 0.1125 moles NaCl
Step 3:
mass NaCl = moles NaCl x MM NaCl = 0.1125 x 58.45 = 6.576 g NaCl
what do you think will happen if there is no coal.
Answer:
If coal and petroleum will get exhausted it will be very difficult for us to transport because most vehicles depends on petroleum, Transport on Earth will became complicated, and if coal will get exhausted we will lose an unique fossil fuel. Coal is used in various domestic and industrial purposes.
Two units of distance used by scientists to describe distances in space are the light-year and astronomical unit. Why do scientists use the light-year instead of Astronomical Units to measure the distances between stars?
The speed of sound is too slow.
Telescopes can measure only in light years
Distant stars can only be seen if there is light
The smaller numbers that result from measuring with larger units are easier to use
Answer:
The smaller numbers that result from measuring with larger units are easier to use .
That’s the one, good luck!
The smaller numbers that result from measuring with larger units are easier to use.
Light year is a unit of astronomical distance that measures the distance traveled by light in one year. The distance between stars can also be measured using parsecs.
The equivalent of this distance in various units is written as follows;
1 light year (ly) = 9.5 x 10¹² km = 63,240 AU1 parsec = 3.26 lyUsing light year to represent the unit of distance is more concise because big measurement can be represented easily with small numbers using this large unit.
Thus, we can conclude that the smaller numbers that result from measuring with larger units are easier to use.
Learn more here:https://brainly.com/question/803764
Fresh milk has a pH of 6.
How do you think the pH
would change as it became
sour? Explain your answer.
How many atoms are found in a 15.5 g sample of bismuth?
Answer:
The number of particles in the 41.8g sample of Bismuth is 12.044 × 10²³
Explanation:
In a 15.5 g sample of bismuth there are 4.47 x 10²² atoms are present.
What do you mean by the term molar mass ?The term molar mass is defined as the ratio between the mass and the amount of substance of any sample of said compound. The molar mass is a property of a substance.
The molar mass of a compound can be calculated by adding the standard atomic masses in g/mol of the constituent atoms.
Given:
Mass of Bismuth = 15.5g
Number of atoms = ?
To find the number atoms, find the number of moles in this element first.
Number of moles = mass/ molar mass
Number of moles = 15.5g / 209
= 0.074mole
1 mole of a substance contains 6.023 x 10²³ atoms
0.074 mole of Bismuth will contain 0.074 x 6.023 x 10²³ atoms
= 4.47 x 10²² atoms
Thus, 4.47 x 10²² atoms are found in a 15.5 g sample of bismuth.
To learn more about the molar mass, follow the link;
https://brainly.com/question/22997914
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What type of chemical reaction is this? CH4 + O2 + CO2 + H20 ?
Answer:
combustion reaction. hope this helps!
1. What is the molar mass of Al2S3?
Answer:
150.17 g/mol
Explanation:
[24 points] A sample of soil has a total volume of 205 cm3. The soil mass when saturated is 361 g. A specific yield test was conducted on the soil by allowing the sample to drain for 24 hours. After drainage the sample mass was 295 g. The soil was then dried and weighed 284 g. What are the specific yield [8 points], specific retention [8 points], and porosity [8 points] of the sample
Answer:
Follows are the solution to the given question:
Explanation:
Dry Soil weight = solid soil weight = [tex]284 \ grams[/tex]
solid soil volume =[tex]205 \ cc[/tex]
saturated mass soil = [tex]361 \ g[/tex]
The weight of the soil after drainage is =[tex]295 \ g[/tex]
Water weight for soil saturation = [tex](361-284) = 77 \ g[/tex]
Water volume required for soil saturation =[tex]\frac{77}{1} = 77 \ cc[/tex]
Sample volume of water: [tex]= \frac{\text{water density}}{\text{water density input}}[/tex]
[tex]= 361- 295 \\\\ = 66 \ cc[/tex]
Soil water retained volume = (draining field weight - dry soil weight)
[tex]= 295 - 284 \\\\ = 11 \ cc.[/tex]
[tex]\text{POROSITY}= \frac{\text{Vehicle volume}}{\text{total volume Soil}}[/tex]
[tex]= \frac{77}{(205 + 77)} \\\\= \frac{77}{(282)} \\\\ = 27.30 \%[/tex]
(Its saturated water volume is equal to the volume of voids)
[tex]\text{YIELD SPECIFIC} = \frac{\text{Soil water volume}}{\text{Soil volume total}}[/tex]
[tex]= \frac{66}{(205+77)}\\\\= \frac{66}{(282)}\\\\=0.2340\\\\ = 0.23[/tex]
[tex]\text{Specific Retention}= \frac{\text{Volume of soil water}}{\text{Total soil volume}}[/tex]
[tex]= \frac{11}{282} \\\\= 0.0390 \\\\ = 0.04[/tex]
