E. Carboxyl. Amino acids contain a carboxyl group, which is an acidic functional group.
This group contains a COOH group, which is composed of a carbon atom double-bonded to an oxygen atom and single-bonded to a hydrogen atom. When in solution, the carboxyl group will donate a hydrogen ion, making it acidicCarboxyl is a functional group in organic chemistry composed of a carbonyl group (C=O) with a hydroxyl group (O-H). It is found in many organic compounds such as amino acids, fatty acids, and carbohydrates. Carboxyl groups are responsible for the acids found in many organic compounds, and they can also act as a source of energy for many metabolic reactions. They are also involved in the formation of hydrogen bonds, which is essential for the formation of proteins and other biological molecules.
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complete question: Amino acids are acids because they always possess which functional group?
A. Phosphate B. Hydroxyl C. Amino D. Carbonyl E. Carboxyl
which gas makes up 78 percent of our atmosphere but can be used by plants only when transformed by bacteria first?
Our atmosphere contains 78 percent nitrogen, but plants can only utilise it once microorganisms have changed it.
Nitrogen (N2) makes up 78% of our atmosphere, but it cannot be used by plants directly. Nitrogen must be transformed into a usable form, such as nitrate (NO3-), by bacteria through the process of nitrogen fixation. The equation for this process is:
N2 + 8H+ + 8e- → 2NH3 + H2
This reaction requires energy, which is usually derived from the oxidation of organic compounds.
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Find the empirical formula of a compound found to contain 26.56% potassium, 35.41% chromium, and the remainder oxygen.
The empirical formula of a compound found to contain 26.56% potassium, 35.41% chromium, and the remainder oxygen is K₂Cr₂O₇.
What is an empirical formula ?The term an Empirical formula is defined as the chemical formula of a compound that gives the ratios of the elements present in the compound but not the actual numbers or arrangement of atoms.
The number of mole of each element in the compound can be identified by dividing each element's percentage with their respective molar weights:
Potassium K = 26.56%
= 26.56 / 39.1
= 0.68
Chromium, Cr = 35.41%
= 35.41/52
= 0.68
Oxygen,
O = 100 - 26.56+35.41
= 38.03/16
= 2.38
Divide each number of moles by the smallest.
K = 0.68/0.68
= 1
Cr = 0.68/0.68
= 1
O = 2.38/0.68
= 3.5
Thus, the empirical formula would be KCrO₃.₅
Multiply all by 2 to remove the fraction:
Then the empirical formula would be K₂Cr₂O₇.
Thus, the empirical formula of a compound found to contain 26.56% potassium, 35.41% chromium, and the remainder oxygen is K₂Cr₂O₇.
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triose phosphate isomerase converts dihydroxyacetone phosphate to glyceraldehyde 3-phosphate through an intermediate with a carbon-carbon double bond ?
Triose phosphate isomerase converts dihydroxyacetone phosphate to glyceraldehyde 3-phosphate through an intermediate with a carbon-carbon double bond is the correction is a) True.
The Dihydroxyacetone phosphate (DHAP) will converted to the glyceradehyde-3-phosphate (G3P) by the enzyme called as the triose phosphate isomerase. this enzyme will catalyzes the isomerization of the three-carbon sugar into the another three-carbon sugar. as the molecular formulas of the DHAP and the G3P are the same, that the reason that they are isomers of the each other.
The trio Phosphate Isomers will catalyzes the transfer of the hydrogen atom from the carbon 1 to other carbon 2.
This question is incomplete, the complete question is :
Triose phosphate isomerase converts dihydroxyacetone phosphate to glyceraldehyde 3-phosphate through an intermediate with a carbon-carbon double bond ?
a) True
b) False
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to what volume should you dilute 131 ml m l of an 7.95 m m cucl2 c u c l 2 solution so that 49.0 ml m l of the diluted solution contains 4.56 g g cucl2 c u c l 2 ?
131 mL of the original volume should be diluted with 131 mL - 49.0 mL = 82.0 mL of water to obtain the desired concentration.
What is volume?
Volume is a measure of the amount of space occupied by a three-dimensional object or substance. It is typically measured in cubic units, such as cubic centimeters (cm^3), cubic meters (m^3), or milliliters (mL).
To solve this problem, we need to first calculate the amount of CuCl2 in 49.0 mL of the original solution, and then determine the volume that should be added to dilute the solution to this concentration.
First, let's calculate the amount of CuCl2 in 49.0 mL of the original solution. Using the concentration of the solution (7.95 mM), we can calculate the number of moles of CuCl2 in 49.0 mL of the solution:
(7.95 mM) * (49.0 mL) * (10^-3 L/mL) = 0.388 moles CuCl2
Next, we can use the moles of CuCl2 to calculate the mass in grams:
0.388 moles * (134.45 g/mol) = 51.99 g
So, 49.0 mL of the original solution contains 51.99 g of CuCl2.
Next, we want to find the volume that should be added to dilute the solution to a concentration such that 49.0 mL of the diluted solution contains 4.56 g of CuCl2. We can use the formula:
C1V1 = C2V2
where C1 is the concentration of the original solution, V1 is the volume of the original solution, C2 is the concentration of the diluted solution, and V2 is the volume of the diluted solution. Rearranging the formula to solve for V2:
V2 = (C1V1) / C2
Plugging in the known values:
V2 = (7.95 mM * 49.0 mL) / (4.56 g / (49.0 mL))
Converting the concentration from mM to g/mL:
V2 = (7.95 mM * 49.0 mL) / (4.56 g / (49.0 mL)) = (7.95 * 10^-3 * 49.0) / (4.56 / 49.0)
V2 = 49.0 mL * (7.95 * 10^-3) / (4.56 / 49.0 g/mL) = 49.0 mL * (7.95 * 10^-3) * (49.0 g/mL) / 4.56 g
V2 = 49.0 mL * 7.95 * 10^-3 * 49.0 / 4.56 = 49.0 mL * 3.52
So, 131 mL of the original volume should be diluted with 131 mL - 49.0 mL = 82.0 mL of water to obtain the desired concentration.
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A student designed an experiment where all metals were first reacted with hot HCl. Is this a good idea?
Reacting metals with hot hydrochloric acid (HCl) can produce hazardous and potentially explosive gas releases, and it can also cause the release of harmful fumes. Therefore, it is not a good idea to use hot hydrochloric acid in this type of experiment without proper safety measures and equipment.
Additionally, the reaction of metals with hydrochloric acid can produce different products depending on the metal, which could make the results of the experiment difficult to interpret or unreliable.
It is important to carefully consider the potential hazards and limitations of any experimental design, and to take appropriate safety precautions when conducting experiments that involve reactive chemicals. Before conducting any experiment, it is advisable to consult with a knowledgeable individual and to thoroughly research the potential risks and safety considerations involved.
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the mass and volume of each sample differ from the mass and volume of the other samples. is it possible for each sample to contain 1 mol of each substance?
Yes, because the number of moles is not dependent on the mass or the volume.
What is Avogadro's law?
Avogadro's law states that "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules". For a given mass of an ideal gas, the volume and amount (moles) of the gas are directly proportional if the temperature and pressure are constant.
Does the number of moles depend on volume?
The volume (V) of an ideal gas varies directly with the number of moles of the gas (n) when the pressure (P) and the temperature (T) are constant.
Does mole depend on mass?
Because of the way in which the mole is defined, for every element, the number of grams in a mole is the same as the number of atomic mass units in the atomic mass of the element.
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which of the following is not correct? 1 mol of o3 has 1.81 x 1024 o atoms 1mol of cl2 has 35.45 g cl 1mol of nacl has 6.022 x 1023 na 1 mole of ca3n2 has 3 moles ca 2
The statement which is not correct is the correct option is 1 mole of Cl₂ has 35.45 g of Cl.
The number of the moles is expressed as :
The number of moles = mass / molar mass
Where,
The moles of the chlorine molecule = 1 mole
The molar mass of the Cl = 35.45 g/mol
The molar mass of the Cl₂ = 2 × 35.45
The molar mass of the Cl₂ = 71 g
The mass of the chlorine = moles × molar mass
The mass of the chlorine = 1 mol × 71 g/mol
The mass of the chlorine = 71 g
Thus the grams of the chlorine molecule present in the 1 mole of the chlorine molecule is 71 g.
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