360 mg / 1000 => 0.36 g
molar mass => 180 /mol
number of moles:
mass of solute / molar mass
0.36 / 180 => 0.002 moles
Volume solution = 200 mL / 1000 => 0.2 L
M = n / V
M = 0.002 / 0.2
M = 0.01 mol/L
hope this helps!
2NH₂ + O₂ → N₂ + 2H₂O
<u>Explanation:</u>
Balancing the equation means, the number of atoms on both sides of the equation must be the same.
In the case of the given equation, we have to find out whether it is balanced or not.
2NH₂ + O₂ → N₂ + 2H₂O
Atoms Number of atoms before balancing after balancing
LHS RHS LHS RHS
N 1 2 2 2
H 2 2 4 4
O 2 1 2 2
To balance the N atoms, we have to put 2 in front of NH₂, and then to balance the H, O atoms, we have to put 2 in front of H₂O, so that each atom in left hand as well as right hand side of the equation was balanced.
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>
