Answer:
18,01528
Explanation:
<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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Answer: Block # 1 , Block # 4 and
Block # 5 will sink.
Block # 2 and 3 will float.
Explanation: The density of water is equal to 1 g/mL. Any density that is less than the density of water will float. Objects with higher density compared to water will eventually sink.
Answer:
It is called the atomic theory I believe.
Explanation:
Answer:
AsF3:C2CI6
4:3
1.3618 moles: 1.02135 moles(1.3618÷4×3)
C2CI6 is the limting reagent
So the number of moles for AsCI3 is 0.817 moles( number of moles of the limting reagant) ÷3 ×4 (according to ratio by balancing chemical equation)=1.09 moles(3 s.f.)
or
Balanced equation
4AsF3 + 3C2Cl6 → 4AsCl3 + 3C2Cl2F4
Use stoichiometry to calculate the moles of AsCl3 that can be produced by each reactant.
Multiply the moles of each reactant by the mole ratio between it and AsCl3 in the balanced equation, so that the moles of the reactant cancel, leaving moles of AsCl3.
Explanation: