The Molar concentration of your analyte solution is 1.17 m
<h3>What is titration reaction?</h3>
- Titration is a chemical analysis procedure that determines the amount of a sample's ingredient by adding a precisely known amount of another substance to the measured sample, with which the desired constituent reacts in a specific, known proportion.
Make use of the titration formula.
The formula is molarity (M) of the acid x volume (V) of the acid = molarity (M) of the base x volume (V) of the base.
if the titrant and analyte have a 1:1 mole ratio. (Molarity is a measure of a solution's concentration represented as the number of moles of solute per litre of solution.)
26 x 1.8 = 40 x M
M = 26 x1.8 /40
M = 1.17
The Molar concentration of your analyte solution is 1.17 m
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Answer: By understanding conversion factors and how they are related to each other
Explanation:
Dimensional Analysis is a step by step approach to solving problems in Physics, Chemistry , and Mathematics. It involves having a clear knowledge and understanding to be able to convert a given unit to another in the same dimension using conversion factors and knowing how they are related to each other.
For instance, In Chemistry, we want to Convert 120mL to L.(note that ml stands for millilitres and ;L stands for litres)
Or first approach will be to write out the conversion factor related to our problem which is
1000ml =1L
such that 120ml = (we cross multiply))
giving us 120ml x 1L/1000ml =0.12L
This same process is applied to convert any type of dimensional analysis problems be it physics or mathematics.
Answer:
Over a vast amount of time, millions of years, the earth gradually cooled. When the temperature dropped enough, water vapor condensed and went from a gas to liquid form. This created clouds. From these clouds, the oceans formed and the oceans absorbed a lot of the carbon dioxide in the atmosphere.
Explanation:
Answer:

Explanation:
Firstly, write the expression for the equilibrium constant of this reaction:
![K_{eq} = \frac{[ADP][Pi]}{ATP}](https://tex.z-dn.net/?f=K_%7Beq%7D%20%3D%20%5Cfrac%7B%5BADP%5D%5BPi%5D%7D%7BATP%7D)
Secondly, we may relate the change in Gibbs free energy to the equilibrium constant using the equation below:

From here, rearrange the equation to solve for K:

Now we know from the initial equation that:
![K_{eq} = \frac{[ADP][Pi]}{ATP}](https://tex.z-dn.net/?f=K_%7Beq%7D%20%3D%20%5Cfrac%7B%5BADP%5D%5BPi%5D%7D%7BATP%7D)
Let's express the ratio of ADP to ATP:
![\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BADP%5D%7D%7B%5BATP%5D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7BK_%7Beq%7D%7D)
Substitute the expression for K:
![\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BADP%5D%7D%7B%5BATP%5D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7BK_%7Beq%7D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7Be%5E%7B-%5Cfrac%7B%5CDelta%20G%5Eo%7D%7BRT%7D%7D%7D)
Now we may use the values given to solve:
![\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}} = [Pi]e^{\frac{\Delta G^o}{RT}} = 1.0 M\cdot e^{\frac{-30 kJ/mol}{2.5 kJ/mol}} = 6.14\cdot 10^{-6}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BADP%5D%7D%7B%5BATP%5D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7BK_%7Beq%7D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7Be%5E%7B-%5Cfrac%7B%5CDelta%20G%5Eo%7D%7BRT%7D%7D%7D%20%3D%20%5BPi%5De%5E%7B%5Cfrac%7B%5CDelta%20G%5Eo%7D%7BRT%7D%7D%20%3D%201.0%20M%5Ccdot%20e%5E%7B%5Cfrac%7B-30%20kJ%2Fmol%7D%7B2.5%20kJ%2Fmol%7D%7D%20%3D%206.14%5Ccdot%2010%5E%7B-6%7D)
The correct answer among the choices is option C. The ion that is part of all nucleic acids is phosphoric acid. Nucleic acids are large biomolecules that is important for all life forms. DNA and RNA are nucleic acids. These biomolecules are made from monomers called nucleotides. Each monomer is composed of 5 carbon sugar, a nitrogeneous base and a phosphate group.