It is tasteless and colorless.
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
Answer:
Percentage error = 1.88 %
Solution:
Data Given:
Mass of Sample = 20.46 g
Volume of Sample = 43.0 mL - 40.0 mL = 3.0 mL
Formula Used:
Density = Mass / Volume
Putting values,
Density = 20.46 g / 3.0 mL
Density = 6.82 g.mL⁻¹
Percentage Error:
Experimental Value = 6.82 g.mL⁻¹
Accepted Value = 6.95 g.mL⁻¹
= 6.82 g.mL⁻¹ / 6.95 g.mL⁻¹ × 100 = 98.12 %
Percentage Error = 100 % - 98.12 %
Percentage error = 1.88 %
Not sure what you are asking. I have two possible answers though...
It could either be more negatively charged, or valence electrons.
The more away from the nucleus a electron is, the more negatively charged it is.
The electrons on the outermost electron shell is valence electrons.
Again, I don't know what you were asking, but one of these answers may be correct.