There are 0.0214 mol Cs in 1.29×10²² atoms Cs.
Use <em>Avogadro’s number</em> to convert atoms of Cs to moles of Cs.
Moles of Cs = 1.29×10²² atoms Cs × (1 mol Cs /6.022 × 10²³ atoms Cs) = 0.0214 mol Cs
<u>We are given:</u>
Mass of NaCl in the given solution = 22.3 grams
Volume of the given solution = 2 L
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<u>Number of Moles of NaCl:</u>
We know that the number of moles = Given mass / Molar mass
Number of moles = 22.3 / 58.44 = 0.382 moles
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<u>Molarity of NaCl in the Given solution:</u>
We know that Molarity of a solution = Moles of Solute / Volume of Solution(in L)
Molarity = 0.382 / 2
Molarity = 0.191 M
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Answer:
At end point there will a transition from pink to colorless.
Explanation:
As the student put the vinegar in the titrator and NaOH in the beaker, it means that he has poured phenolphthalein in the NaOH solution.
The pH range of phenolphthalein is 8.3-10 (approx), it means it will show pink color in basic medium.
So on addition of phenolphthalein in NaOH the solution will become pink in color.
When we start pouring vinegar from titrator neutralization of NaOH will begin.
On complete neutralization , on addition of single drop of vinegar the solution will become acidic and there will be complete disappearance of pink color solution in the beaker.
T<span>his is a straightforward question related to the surface energy of the droplet. </span>
<span>You know the surface area of a sphere is 4π r² and its volume is (4/3) π r³. </span>
<span>With a diameter of 1.4 mm you have an original droplet with a radius of 0.7 mm so the surface area is roughly 6.16 mm² (0.00000616 m²) and the volume is roughly 1.438 mm³. </span>
<span>The total surface energy of the original droplet is 0.00000616 * 72 ~ 0.00044 mJ </span>
<span>The five smaller droplets need to have the same volume as the original. Therefore </span>
<span>5 V = 1.438 mm³ so the volume of one of the smaller spheres is 1.438/5 = 0.287 mm³. </span>
<span>Since this smaller volume still has the volume (4/3) π r³ then r = cube_root(0.287/(4/3) π) = cube_root(4.39) = 0.4 mm. </span>
<span>Each of the smaller droplets has a surface area of 4π r² = 2 mm² or 0.0000002 m². </span>
<span>The surface energy of the 5 smaller droplets is then 5 * 0.000002 * 72.0 = 0.00072 mJ </span>
<span>From this radius the surface energy of all smaller droplets is 0.00072 and the difference in energy is 0.00072- 0.00044 mJ = 0.00028 mJ. </span>
<span>Therefore you need roughly 0.00028 mJ or 0.28 µJ of energy to change a spherical droplet of water of diameter 1.4 mm into 5 identical smaller droplets. </span>
