Lets assume x volume of NaOH and x volume of HCl are added together.
NaOH ---> Na⁺ + OH⁻
NaOH is a strong base therefore it completely ionizes and releases OH⁻ ions into the medium
HCl ---> H⁺ + Cl⁻
HCl is a strong base and completely ionizes and releases H⁺ ions in to the medium. number of NaOH moles in 1 L - 0.1 mol
Therefore in x L - 0.1 /1 * x = 0.1x moles of NaOH present
Similarly in HCl x L contains - 0.1x moles of HCl
H⁺ + OH⁻ ---> H₂O
Due to complete ionisation, 0.1x moles of H⁺ ions and 0.1x moles of OH⁻ ions react to form 0.1x moles of H₂O. Therefore all H⁺ and OH⁻are completely used up and yield water molecules.
Then at this point the H⁺ and OH⁻ ions in the medium come from the weak dissociation of water. This is equivalent to 1 x 10⁻⁷M
pH = -log [H⁺]
pH = -log [10⁻⁷]
pH = 7
pH is therefore equals to 7 which means the solution is neutral
Answer: Cardiovascular System
Explanation:
This involves your heart, blood, veins, and arteries
Answer:
particles in 2 moles.
Explanation:
The number of particles that are contained in one mole, the international unit of amount of substance: by definition, exactly 6.022×10²³, and it is dimensionless. It is named after the scientist Amedeo Avogadro.
It is also known as Avogadro's constant.
∴ Number of particles in one mole = 
∴ Number of particles in 2 mole = 2 times Number of particles in one mole
∴ Number of particles in 2 mole=
Hence there are particles in 2 moles.
Answer:
Uracil
Explanation:
The base that will NOT combine with 2-deoxyribose to form a nucleic acid is Uracil.
2-deoxyribose is a pentose sugar found in the DNA (Deoxyribonucleic acid). It is devoid of oxygen in its 2' position. The bases found in DNA are Adenine, Guanine, Cytosine and Thymine. Adenine, Guanine, and Cytosine are also found in RNA (Ribonucleic acid). Thymine is not present in RNA, it is only found in DNA. The base found in RNA is Uracil which in turn is not present in DNA. The five carbon sugar present in RNA is ribose which combines with Uracil.
Answer: Concentration of 
in the equilibrium mixture is 0.31 M
Explanation:
Equilibrium concentration of 
= 0.729 M
The given balanced equilibrium reaction is,
Initial conc. x 0 0
At eqm. conc. (x-2y) M (y) M (3y) M
The expression for equilibrium constant for this reaction will be:
3y = 0.729 M
y = 0.243 M
![K_c=\frac{[y]\times [3y]^3}{[x-2y]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5By%5D%5Ctimes%20%5B3y%5D%5E3%7D%7B%5Bx-2y%5D%5E2%7D)
Now put all the given values in this expression, we get :
concentration of in the equilibrium mixture = 
Thus concentration of in the equilibrium mixture is 0.31 M
