What are the components of a nucleotide
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This problem is providing us with the mass of hydrochloric acid and the volume of solution and asks for the pH of the resulting solution, which turns out to be 1.477.
<h3>pH calculations</h3>In chemistry, one can calculate the pH of a solution by firstly obtaining its molarity as the division of the moles of solute by the liters of solution, so in this case for HCl we have:
Next, due to the fact that hydrochloric acid is a strong acid, we realize its concentration is nearly the same to the released hydrogen ions to the solution upon ionization. Thereby, the resulting pH is:
Which conserves as much decimals as significant figures in the molarity.
Learn more about pH calculations: brainly.com/question/1195974
Answer:
The amount of heat to absorb is 6,261 J
Explanation:
Calorimetry is in charge of measuring the amount of heat generated or lost in certain physical or chemical processes.
The total energy required is the sum of the energy to heat the ice from -20 ° C to ice of 0 ° C, melting the ice of 0 ° C in 0 ° C water and finally heating the water to 60 ° C.
So:
- Heat required to raise the temperature of ice from -20 °C to 0 °C
Being the sensible heat of a body the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous), the expression is used:
Q = c * m * ΔT
Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation (ΔT=Tfinal - Tinitial).
In this case, m= 10 g, specific heat of the ice= 2.1 and ΔT=0 C - (-20 C)= 20 C
Replacing: Q= 10 g*2.1 *20 C and solving: Q=420 J
- Heat required to convert 0 °C ice to 0 °C water
The heat Q necessary to melt a substance depends on its mass m and on the called latent heat of fusion of each substance:
Q= m* ΔHfusion
In this case, being 1 mol of water= 18 grams: Q= 10 g*= 3.333 kJ= 3,333 J (being kJ=1,000 J)
- Heat required to raise the temperature of water from 0 °C to 60 °C
In this case the expression used in the first step is used, but being: m= 10 g, specific heat of the water= 4.18 and ΔT=60 C - (0 C)= 60 C
Replacing: Q= 10 g*4.18 *60 C and solving: Q=2,508 J
Finally, Qtotal= 420 J + 3,333 J + 2,508 J
Qtotal= 6,261 J
<u><em> The amount of heat to absorb is 6,261 J</em></u>
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