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AlekseyPX
3 years ago
11

PLEASE HELP, THIS IS TIMED!!!!!!!!!!!!!!!!!!!!!!!!!!Convert 22.6g AgNO3 to moles.

Chemistry
1 answer:
andreev551 [17]3 years ago
5 0

Answer:

the answer b because

The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles AgNO3, or 169.8731 grams.

Explanation:

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The vapor pressure of water is 23.76 mm Hg at 25 °C. A nonvolatile, nonelectrolyte that dissolves in water is sucrose. Calculate
MrRissso [65]

Answer : The vapor pressure of solution is 23.67 mmHg.

Solution:

As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.

The formula for relative lowering of vapor pressure will be,

\frac{p^o-p_s}{p^o}=\frac{w_2M_1}{w_1M_2}

where,

p^o = vapor pressure of pure solvent  (water) = 23.76 mmHg

p_s = vapor pressure of solution= ?

w_2 = mass of solute  (sucrose) = 12.25 g

w_1 = mass of solvent  (water) = 176.3 g

M_1 = molar mass of solvent (water) = 18.02 g/mole

M_2 = molar mass of solute (sucrose) = 342.3 g/mole

Now put all the given values in this formula ,we get the vapor pressure of the solution.

\frac{23.76-p_s}{23.76}=\frac{12.25\times 18.02}{176.3\times 342.3}

p_s=23.67mmHg

Therefore, the vapor pressure of solution is 23.67 mmHg.

7 0
3 years ago
What is the energy required to change a spherical drop of water to five smaller spherical drops of equal size? At room temperatu
pav-90 [236]
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>

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<span>The five smaller droplets need to have the same volume as the original. Therefore </span>

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<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>
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