<span> 52.0ml of 0.35M CH3COOH : 0.052 L(0.35M) = .0182 mol of CH3COOH.
</span>
<span>31.0ml of 0.40M NaOH : .031 L(0.40M) = .0124 mol of NaOH.
</span>
<span>After the reaction, .0124 Mol CH3COO- is generated and .058 mol CH3COOH is left un-reacted. The concentration would be 12.4/V and 5.8/V, respectively. Therefore:
</span>
<span>pH = -log([H+]) = -log(Ka*[CH3COOH]/[CH3COO-]) </span>
<span>= -log(1.8x10^-5*5.8/12.4) = 5.07</span>
Answer:
42.9
∘
C
Explanation:
The idea here is that the problem is providing you with the amount of heat needed to raise the temperature of a given sample of water from an initial temperature to its boiling point, i.e. to 100
Imagine a chemist is in the lab and trying to make some chemical reactions happen. In one reaction she reacts chemicals in an exothermic reaction and there is an increase in entropy. A second chemical reaction she is trying to run is endothermic and there is a decrease in entropy. Which of the two reactions is more likely to occur and why?
Answer and Explanation:
Because metallic bonding involves delocalized electrons. It is described as a "<em>sea of electrons</em>", because the electrons are not confined around the nucleus of metal atoms, but they are delocalized: thay can be located in one nucleus and then in another neighbor atom. Thus, the electrons have more freedom to move from one part of the metal to another and electricity is well conducted.
Answer:
14.57g
Explanation:
Given parameters:
Mass of dish + ball = 15.6g
Initial volume of water in the cylinder = 26.7mL
Final volume of water in the cylinder = 38.9mL
Mass of dish = ?
Unknown
Mass of the ball = ?
Solution;
Since the mass of ball and dish is 15.6g,
Mass of the ball =Mass of ball + dish - mass of the dish
Insert the parameters and solve;
Mass of the ball = 15.6g - 1.03g = 14.57g
