Answer:
Molarity = 1.57 M
Explanation:
Given data:
Volume of solution = 350 mL (350 mL × 1L /1000 ml = 0.35 L)
Mass of NaCl = 32.0 g
Molarity of solution = ?
Solution:
Number of moles of solute:
Number of moles = mass/molar mas
Number of moles = 32.0 g/ 58.44 g/mol
Number of moles = 0.55 mol
Molarity:
Molarity = number of moles / volume in L
Molarity = 0.55 mol / 0.35 mL
Molarity = 1.57 mol/L
Molarity = 1.57 M
Acidity I believe...sorry if I'm wrong just got on the subject
It turns chemical into electricity
Answer: After a few aspirin tablets have been swallowed, the concentration of acetylsalicylic acid in the stomach is 0.20 M. Calculate the percent ionization of the acid under these conditions. Therefore, the percent ionization is - X 100% = 3.8%.
Answer: 1.14
Explanation:
![HCl+NaOH\rightarrow NaCl+H_2O](https://tex.z-dn.net/?f=HCl%2BNaOH%5Crightarrow%20NaCl%2BH_2O)
To calculate the molarity of acid, we use the equation given by neutralization reaction:
![n_1M_1V_1=n_2M_2V_2](https://tex.z-dn.net/?f=n_1M_1V_1%3Dn_2M_2V_2)
where,
are the n-factor, molarity and volume of acid which is ![HCl](https://tex.z-dn.net/?f=HCl)
are the n-factor, molarity and volume of base which is NaOH.
We are given:
![n_1=1\\M_1=?\\V_1=10.0mL\\n_2=1\\M_2=0.1M\\V_2=7.2mL](https://tex.z-dn.net/?f=n_1%3D1%5C%5CM_1%3D%3F%5C%5CV_1%3D10.0mL%5C%5Cn_2%3D1%5C%5CM_2%3D0.1M%5C%5CV_2%3D7.2mL)
Putting values in above equation, we get:
![1\times M_1\times 10.0=1\times 0.1\times 7.2\\\\M_1=0.072M](https://tex.z-dn.net/?f=1%5Ctimes%20M_1%5Ctimes%2010.0%3D1%5Ctimes%200.1%5Ctimes%207.2%5C%5C%5C%5CM_1%3D0.072M)
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)
![pH=-log(0.072)=1.14](https://tex.z-dn.net/?f=pH%3D-log%280.072%29%3D1.14)
Thus the pH of the gastric juice is 1.14