<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:
There are 3 steps of this problem.
Explanation:
Step 1.
Wet steam at 1100 kPa expands at constant enthalpy to 101.33 kPa, where its temperature is 105°C.
Step 2.
Enthalpy of saturated liquid Haq = 781.124 J/g
Enthalpy of saturated vapour Hvap = 2779.7 J/g
Enthalpy of steam at 101.33 kPa and 105°C is H2= 2686.1 J/g
Step 3.
In constant enthalpy process, H1=H2 which means inlet enthalpy is equal to outlet enthalpy
So, H1=H2
H2= (1-x)Haq+XHvap.........1
Putting the values in 1
2686.1(J/g) = {(1-x)x 781.124(J/g)} + {X x 2779.7 (J/g)}
= 781.124 (J/g) - x781.124 (J/g) = x2779.7 (J/g)
1904.976 (J/g) = x1998.576 (J/g)
x = 1904.976 (J/g)/1998.576 (J/g)
x = 0.953
So, the quality of the wet steam is 0.953
Answer:
why did you post a link to brainly when we are on brainly already??????
Explanation:
True.....................................
The new volume of the bag will be 789.5 mL.
<u>Explanation:</u>
As per the Charles law, at constant pressure the volume of the gas is directly related to its temperature in Kelvin (K). That is as the temperature increases, the gas expands and vice-versa.
V1 = 250 ml
V2 = ?
T1 = 19° C
T2 = 60° C
Now we have to rewrite the equation to get the new volume as,
V2 = 
=
<em> </em>= 789.47 ≈ 789.5 ml.
So the new volume of the bag will be 789.5 mL.
