Answer:
47.5 mL
Solving:
M1 = 4.00 M
V1 = ?
M2 = 0.760 M
V2 = 0.250 L
---
M1 * V1 = M2 * V2
V1 = ( M2 * V2 ) / M1
V1 = ( 0.760 * 0.250 ) / 4.00
V1 = ( 0.190 ) / 4.00
V1 = 0.0475 L
2 resonance structure.
O=S⁺-O⁻ and O⁻-S⁺=O, angles between S and O are 120°, bond order is 1,5.
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
Answer:
subatomic particles are smaller then atoms