Answer:
A small amount of solute dissolved in a larger amount of solvent.
Take this analogy to help you understand, if you were to put a teaspoon of sugar in a liter of water it would dissolve, but if you put a sack of sugar in it it would not dissolve! The solute is what is being dissolved and the solvent is what dissolves the solute, so that eliminates some of the options.
Answer:the energy possessed by a body by virtue of its position relative to others, stresses within itself, electric charge, and other factors.
Explanation:
41.38 % Mg
55.17 % O
3.45 % H
Explanation:
What is the percent composition of magnesium hydroxide Mg(OH)₂?
To find the percent composition we follow the next algorithm.
First we calculate the molar mass of Mg(OH)₂:
molar mass of Mg(OH)₂ = molar mass of Mg × 1 + molar mass of O × 2 + molar mass of H × 2
molar mass of Mg(OH)₂ = 24 × 1 + 16 × 2 + 1 × 2 = 58 g/mole
Now we devise the next reasoning:
if in 58 g of Mg(OH)₂ there are 24 g of Mg, 32 g of O and 2 g of H
then in 100 g of Mg(OH)₂ there are X g of Mg, Y g of O and Z g of H
X = (100 × 24) / 58 = 41.38 % Mg
X = (100 × 32) / 58 = 55.17 % O
X = (100 × 2) / 58 = 3.45 % H
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percent composition
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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>
