Answer: 1.14
Explanation:
To calculate the molarity of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is 
are the n-factor, molarity and volume of base which is NaOH.
We are given:
Putting values in above equation, we get:
To calculate pH of gastric juice:
molarity of 
= 0.072
![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
Thus the pH of the gastric juice is 1.14
Properties of a solution that depend only on the ratio of the number of particles of solute and solvent in the solution are known as colligative properties. For this problem, we use boiling point elevation concept.
ΔT(boiling point) = (Kb)mi
ΔT(boiling point) = (0.51 C-kg / mol )(4.0 mol / 2.05 kg ) (2)
ΔT(boiling point) = 1.99 C
T(boiling point) = 101.99 C
Answer:
The six member ring and the position of the -OH group on the carbon (#4) identifies glucose from the -OH on C # 4 in a down projection in the Haworth structure). Fructose is recognized by having a five member ring and having six carbons, a hexose.
Answer:
It is a sample of matter with both constant and definite composition
Explanation:
Answer:
CHO
Explanation:
Carbon = 41%, Hydrogen = 4.58%, oxygen = 54.6%
Step 1:
Divide through by their respective relative atomic masses
41/ 12, 4.58/1, 54.6/16
3.41 4.58 3.41
Step 2:
Divide by the lowest ratio:
3.41/3.41, 4.58/3.41, 3.41/3.41
1, 1, 1
Hence the empirical formula is CHO
