The correct question is as follows: 0.500 moles of potassium oxide is dissolved in enough water to make 2.00 L of solution. Calculate the molarity of this solution (plz help!)

Answer: The molarity of this solution is 0.25 M.

Explanation:

Molarity is the number of moles of a substance divided by volume in liter.

As it is given that there are 0.5 moles of potassium oxide in 2.00 L of water so, the molarity of this solution is calculated as follows.

$Molarity = \frac{moles}{volume(in L)}\\= \frac{0.5 moles}{2.00 L}\\= 0.25 M$

Thus, we can conclude that molarity of this solution is 0.25 M.

4 0

3 years ago

im making a poster for chemistry. the topic is acid and base. i have to make a creative title to go with the poster. any ideas?

REY [17]

The correct answer for the question that is being presented above is this one: Right now, I haven't seen the poster but there are things that i have in mind. It can be "The Game of Acids and Bases," or you can another one like, "Unity Amidst Diversity."

3 0

3 years ago

Which combination may be used to prepare a buffer having a ph of 8. 8?

7nadin3 [17]

The combination used in the preparation of a buffer with a pH around 8.8 accounts for ka = 7 * 10 -3 for h 3po4

The ph of the buffer can be shown as:

pH = pKa + log [Salt] /[ Acid ]

[Salt] /[ Acid ] = x

For h3po4 with ka= 7 × 10–3

8.8 = - log (7 × 10^–3) + log x

8.8 = 2.21 + log x

Thus, the value of log x is coming positive and therefore can be used for preparing buffer.

For h2po4- with ka= 8 × 10–8

8.8 = - log (8 × 10^–8) + log x

8.8 = 7.14 + log x

Thus, the value of log x is coming positive and therefore can be used for preparing buffer.

For hpo42– with ka= 5 × 10–13

8.8 = - log (5 × 10–13) + log x

8.8 = 12.31 + log x

Thus, the value of log x is coming negative and therefore can not be used for preparing buffer.

Hence, the correct answer is option A

Learn more about buffering systems here,

brainly.com/question/16556401

# SPJ4

4 0

2 years ago