THE ANSWER IS B. The reducing of oxidation state (-1 to -2, a change of -1, a REDUCTION) is what makes it a reduction.
Answer:
One can determine the specific heat of the metal through using the clarimeter, water, thermometer and using heat equations.
Explanation:
You can learn about heat effects and calorimetery through a simple experiment by boiling water and heating up the metal in it. Then, pour it into your calorimeter and the heat will flow from the metal to the water. The two equlibria will meet: the metal will loose heat into its surroundings (the water) and teh water will absorb the heat. The heat flow for the water is the same as it is for the metal, the only difference being is the negative sign indicating the loss of the heat of the metal.
In terms of theromdynamics, we can deteremine the heat flow for the metal becasue it would be equal to the mangnitued but opposite in direction. Thus, we can say that the specific heat of water qH2O = -qmetal.
Answer:
Explanation:
Hello there!
In this case, according to the Henderson-Hasselbach equation, it is possible to write:
![pH=pKa+log(\frac{[A^-]}{[HA]} )](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D%20%29)
Next, since we are given the pH and the [A–]/[HA] ratio, we can solve for the pKa as shown below:
![pKa=pH-log(\frac{[A^-]}{[HA]} )](https://tex.z-dn.net/?f=pKa%3DpH-log%28%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D%20%29)
Now, we plug in the values to obtain:
Next, Ka is:
Best regards!
Answer:
Chloroplast
Explanation:
Chloroplasts are found in plant cells and are a shade of green, they capture light energy and use it to make energy!
hope i helped you out:)
