Mass of sodium is 23 and mass of fluirine is 19 there mass of NaF is 42
42 g = 1 mole therefor 4.5 moles will have
4.5 × 42 = 189 g
Answer: There are 0.5 grams of barium sulfate are present in 250 of 2.0 M 
solution.
Explanation:
Given: Molarity of solution = 2.0 M
Volume of solution = 250 mL
Convert mL int L as follows.
Molarity is the number of moles of solute present in liter of solution. Hence, molarity of the given solution is as follows.
Thus, we can conclude that there are 0.5 grams of barium sulfate are present in 250 of 2.0 M solution.
Shred red cabbage ~ (3/4 of a very small head)
Put the cabbage pieces in a small container ~ ( you can use a Pyrex-4-cup measure, a bowl or even a plastic zipper bag)
Cover the cabbage with very hot water. Let it sleep until the water has cooled. (somewhere between lukewarm and room-temperature)
The purple liquid you've made is your indicator.
Pour it into a container and compost the cabbage.
Now look for substances that may be acids or bases.
Liquids are good, like fruits.
You can also use solids around for baking are good too. (such as baking soda, salt, sugar, cream of tartar...)
Get containers for mixing (such as tea cups, because they are small, shallow and white inside)
Pour the indicator into the tea cups and add an acid or base.
Lemon juice, rice wine vinegar, and apple cider vinegar, turn the cabbage-water indicator into a pink.
Orange juice or fresh oranges (same thing) turn the cabbage-water indicator into an orangish-pinkish color.
Baking soda turns the cabbage-water indicator blue.
Milk (non-fat) turns the cabbage-water indicator turn opaque and milky, yet purple.
An egg white (which won't get into the solution immediately until after a lot of stirring) turns the cabbage-water indicator blue.
Hint:
Bases mostly turn the indicator towards blue-ish colors such as purple, light blue, dark blue, opaque blue...
Acids mostly turn the indicator towards pink-ish colours such as orange-ish pink, floral pink...
(You'll have to keep on testing the cabbage-water indicator in after a day or two to see if the indicator quality persists or degrades.
So in your question that ask to calculate the Ph result of the resulting solution if 26 ml of 0.260 M HCI(aq) is added to the following substance. The the result are the following:
A. The result is pH= 14-pOH
B. There are 10ml of 0.26m HCL excees in this reaction so the answer is log(H)+
