Answer:
Nucleotides
Explanation:
Nucleotides are the organic molecules which serve as monomer units for the formation of nucleic acid polymers which are the deoxyribonucleic acid and the ribonucleic acid (RNA) and both are the essential biomolecules within the life on the Earth.
Nucleotides are building blocks of the nucleic acids. They are the molecules which are composed of three sub units which are:
- Nitrogenous base which is also called as nucleobase
- Five-carbon sugar which can be ribose or deoxyribose
- At least one phosphate group which is attached to the sugar.
<span>Answer:
For this problem, you would need to know the specific heat of water, that is, the amount of energy required to raise the temperature of 1 g of water by 1 degree C. The formula is q = c X m X delta T, where q is the specific heat of water, m is the mass and delta T is the change in temperature. If we look up the specific heat of water, we find it is 4.184 J/(g X degree C). The temperature of the water went up 20 degrees.
4.184 x 713 x 20.0 = 59700 J to 3 significant digits, or 59.7 kJ.
Now, that is the energy to form B2O3 from 1 gram of boron. If we want kJ/mole, we need to do a little more work.
To find the number of moles of Boron contained in 1 gram, we need to know the gram atomic mass of Boron, which is 10.811. Dividing 1 gram of boron by 10.811 gives us .0925 moles of boron. Since it takes 2 moles of boron to make 1 mole B2O3, we would divide the number of moles of boron by two to get the number of moles of B2O3.
.0925/2 = .0462 moles...so you would divide the energy in KJ by the number of moles to get KJ/mole. 59.7/.0462 = 1290 KJ/mole.</span>
Answer:
a) Unsaturated
b) Supersaturated
c) Unsaturated
Explanation:
A saturated solution contains the <u>maximum amount of a solute that will dissolve in a given solvent at a specific temperature</u>.
An unsaturated solution contains <u>less solute than it has the capacity to dissolve. </u>
A supersaturated solution, <u>contains more solute than is present in a saturated solution</u>. Supersaturated solutions are not very stable. In time, some of the solute will come out of a supersaturated solution as crystals.
According to these definitions and considering that the solubility of KCl in 100 mL of H₂O at <u>20 °C is 34 g</u>, and at <u>50 °C is 43 g</u> we can label the solutions:
a) 30 g in 100 mL of H₂O at 20 °C ⇒ unsaturated
b) 65 g in 100 mL of H₂O at 50 °C ⇒ supersaturated
c) 42 g in 100 mL of H₂O at 50 °C and slowly cooling to 20 °C to give a clear solution <u>with no precipitate</u> ⇒ unsaturated (if it were saturated it would have had precipitate)
Since
potassium and phosphate is what we are to find for and they are both found in
the potassium phosphate solution, therefore we solve for this one first on the
basis of the phosphate.
The formula
for finding the volume given the concentration and number of moles is:
Volume =
number of moles / concentration in Molarity
Volume
potassium phosphate required = 30 mmol phosphate / (3 mmol / mL)
<u>Volume
potassium phosphate required = 10 mL</u>
This would
also contain potassium in amounts of:
Amount of
potassium in potassium phosphate = 10 mL (4.4 meg / mL)
Amount of
potassium in potassium phosphate = 44 meg
Therefore
the potassium chloride required is:
Volume of
potassium chloride = (80 meg – 44 meg) / (2 meg / mL)
<span><u>Volume of
potassium chloride = 72 mL</u></span>
Answer:
A. Solution, Colloid, Suspension
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