Answer:
0.73L
Explanation:
The following data were obtained from the question :
V1 = 0.65 L
P1 = 3.4 atm
T1 = 19°C = 19 + 273 = 292K
V2 =?
P2 = 3.2 atm
T2 = 36°C = 36 + 273 = 309K
The bubble's volume near the top can be obtain as follows:
P1V1 /T1 = P2V2 /T2
3.4 x 0.65/292 = 3.2 x V2 /309
Cross multiply to express in linear form as shown below:
292 x 3.2 x V2 = 3.4 x 0.65 x 309
Divide both side by 292 x 3.2
V2 = (3.4 x 0.65 x 309) /(292 x 3.2)
V2 = 0.73L
Therefore, the bubble's volume near the top is 0.73L
Given that <span>sample a has a higher melting point than sample
b. Therefore, sample a is a longer chain of a </span><span>fatlike solid substance. It could also be that the bonds present in sample a is much stronger which will require more energy to break. Hope this answers the question.</span>
This is a straightforward question related to the surface energy of the droplet.
<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>
Answer:
the ion present in the original solution is Ca2+
Explanation:
Precipitation reactions occur when cations and anions in aqueous solution combine to form an insoluble ionic solid called a precipitate.
<u>Step1</u> : If we add Nacl to the solution, there is no precipitate formed
⇒The only possible ion that can form a precipate with Cl- is Ag+; since there is no precipitate formed, Ag+ is not present
<u>Step2</u> : If we add Na2SO4 to the solution, a white precipitate is formed
The possible ions to bind at SO42- are Ca2+ and Fe2+
But the white precipitate formed, points in the direction of Ca2+
⇒This means calcium is present
<u>Step3</u> : If we add Na2CO3 to the filtered solution, there is a precipate formed
Ca2+ will bind also with CO32- and form a precipitate
So the ion present in the original solution is Ca2+