Answer:
A
Explanation:
Ball bounces because of the difference in the air pressure on the outside of the ball, and opposite the point of impact, and the pressure inside of the ball. This means that the inside of the ball now has less space to contain the air molecules contained inside the ball, increasing the pressure inside the ball.
Answer:
237.2 mL.
Explanation:
- We have the rule: at neutralization, the no. of millimoles of acid is equal to the no. of millimoles of the base.
(XMV) acid = (XMV) base.
where, X is the no. of (H) or (OH) reproducible in acid or base, respectively.
M is the molarity of the acid or base.
V is the volume of the acid or base.
<em>(XMV) HCl = (XMV) NaOH.</em>
<em></em>
For HCl; X = 1, M = 0.5 M, V = ??? mL.
For NaOH, X = 1, M = 1.54 M, V = 77.0 mL.
<em>∴ V of HCl = (XMV) NaOH / (XV) HCl = (</em>1)(1.54 M)(77.0 mL) / (1)(0.5 M) = <em>237.2 mL.</em>
Insect's or Bugs preserved in Amber
Answer:
Explanation:
Hello,
In this case, for a concentration of 0.42 M of benzoic acid whose Ka is 6.3x10⁻⁵ in 0.33 M sodium benzoate, we use the Henderson-Hasselbach equation to compute the required pH:
![pH=pKa+log(\frac{[base]}{[acid]} )](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29)
Whereas the concentration of the base is 0.33 M and the concentration of the acid is 0.42 M, thereby, we obtain:
![pH=-log(Ka)+log(\frac{[base]}{[acid]} )\\\\pH=-log(6.3x10^{-5})+log(\frac{0.33M}{0.42M} )\\\\pH=4.1](https://tex.z-dn.net/?f=pH%3D-log%28Ka%29%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29%5C%5C%5C%5CpH%3D-log%286.3x10%5E%7B-5%7D%29%2Blog%28%5Cfrac%7B0.33M%7D%7B0.42M%7D%20%29%5C%5C%5C%5CpH%3D4.1)
Regards.
