Answer:
Carbon dioxide reacts with limewater (a solution of calcium hydroxide, Ca(OH) 2), to form a white precipitate (appears milky) of calcium carbonate, CaCO 3.
Hope it helps you! :)
First, we have to remember the molarity formula:

Part 1:
In this case, our solute is sodium nitrate (NaNO3), and we have the mass dissolved in water, then we have to convert grams to moles. For that, we need the molecular weight:

Then, we calculate the moles present in the solution:

Now, we have the necessary data to calculate the molarity (with the solution volume of 200 mL):

The molarity of this solution equals 0.2339 M.
Part 2:
In this case, we have the same amount (in moles and mass) of sodium nitrate, but a different volume of solution, then we only have to change it:

So, the molarity of this solution is 0.1701 M.
The correct answer is option B. X-rays are the waves found in the electromagnetic spectrum at a wavelength of 1x10^-11 - 1x10^-8 m and frequencies of 3x10^16 to 3x10^19 Hz. These waves penetrate easily through certain materials which made them usefule in the field of medicine.
Answer:
cause they like it that way
Explanation:
btw cant tell if this is legit or not
<u>36 ml of NaOh and</u><u> 464 ml</u><u> of </u><u>HCOOH</u><u> would be enough to form 500 ml of a buffer with the same pH as the buffer made with </u><u>benzoic acid </u><u>and NaOH.</u>
What is benzoic acid found in?
- Some natural sources of benzoic acid include: Fruits: Apricots, prunes, berries, cranberries, peaches, kiwi, bananas, watermelon, pineapple, oranges.
- Spices: Cinnamon, cloves, allspice, cayenne pepper, mustard seeds, thyme, turmeric, coriander.
Calculate the amount of moles in NaOH and benzoic acid. This calculation is done by multiplying molarity by volume.
Amount of moles of NaOH -2 × 0.025 = 0.05 mol
Amount of moles of benzoic acid 2 × 0.475 = 0.095 mol
In this case, we can calculate the pH produced by the buffer of these two reagents, as follows


We must repeat this calculation, with the values shown for HCOOH and NaOH. In this case, we can calculate as follows




Now we must solve the equation above. This will be done using the following values

With these values, we can calculate the volumes of NaOH and HCOOH needed to make the buffer.
NaOH volume
( 0.5 - 0.464)L
0.036L .................... 36ml
HCOOH volume
500 - 36 = 464mL
Learn more about benzoic acid
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