Answer:
Explanation:
Use your other senses such as touch or smell, does it react in different temperatures, is it then visible under a microscope, does it react with different elements or gases? Think outside the box.
Answer:
C.) The number after it is between 5 and 9
Explanation:
If the digit to the right is less than five, you must leave the digit you want to round to as it is.
But, if the digit to the right is greater than or equal to five, you must increase the value of the digit you're rounding to by one. And any remaining digits before the decimal point become zeros, and any that are after the decimal point are dropped.
Hope this helps you out! : )
Answer:
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to solve for the molarity of the KOH solution by knowing this base react in a 1:1 mole ratio with nitric acid, HNO3; thus, we can write the following equation, as their moles are the same at the endpoint:
Which in terms of molarities and volumes is:
Thus, we solve for the molarity of the base (KOH) to obtain:
Regards!
Answer:
151200 minutes.
Explanation:
From the question given above, the following data were obtained:
Time (in week) = 15 weeks
Time (in min) =?
Next, we shall convert 15 weeks to days. This can be obtained as follow:
1 week = 7 days
Therefore,
15 weeks = 15 weeks × 7 days / 1 week
15 weeks = 105 days
Next, we shall convert 105 days to hours. This can be obtained as follow:
1 day = 24 h
Therefore,
105 days = 105 days × 24 h / 1 day
105 days = 2520 h
Finally, we shall convert 2520 h to mins. This can be obtained as follow:
1 h = 60 mins
Therefore,
2520 h = 2520 h × 60 mins / 1 h
2520 h = 151200 mins
Thus, 15 weeks is equivalent to 151200 minutes.
Answer:
d. 0.121 M HC2H3O2 and 0.116 M NaC2H3O2
Explanation:
Hello,
In this case, since the pH variation is analyzed via the Henderson-Hasselbach equation:
![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)
We can infer that the nearer to 1 the ratio of of the concentration of the base to the concentration of the acid the better the buffering capacity. In such a way, since the sodium acetate is acting as the base and the acetic acid as the acid, we have:
a. ![\frac{[Base]}{[Acid]}=\frac{0.497M}{0.365M}=1.36](https://tex.z-dn.net/?f=%5Cfrac%7B%5BBase%5D%7D%7B%5BAcid%5D%7D%3D%5Cfrac%7B0.497M%7D%7B0.365M%7D%3D1.36)
b. ![\frac{[Base]}{[Acid]}=\frac{0.217M}{0.521M}=0.417](https://tex.z-dn.net/?f=%5Cfrac%7B%5BBase%5D%7D%7B%5BAcid%5D%7D%3D%5Cfrac%7B0.217M%7D%7B0.521M%7D%3D0.417)
c. ![\frac{[Base]}{[Acid]}=\frac{0.713M}{0.821M}=0.868](https://tex.z-dn.net/?f=%5Cfrac%7B%5BBase%5D%7D%7B%5BAcid%5D%7D%3D%5Cfrac%7B0.713M%7D%7B0.821M%7D%3D0.868)
d. ![\frac{[Base]}{[Acid]}=\frac{0.116M}{0.121M}=0.959](https://tex.z-dn.net/?f=%5Cfrac%7B%5BBase%5D%7D%7B%5BAcid%5D%7D%3D%5Cfrac%7B0.116M%7D%7B0.121M%7D%3D0.959)
Therefore, the d. solution has the best buffering capacity.
Regards.
