If the concentration of water inside a cell is higher than the concentration of water outside a cell, osmosis will take place, as water will move from an area of low solute concentration inside the cell to higher solute concentration, outside the cell.
Answer:
15.69 dozen
Explanation:
Mass of penny = 5 g
Dozens of penny =..?
Next, we shall convert 5 g to gross. This can be obtained as follow:
3824 g = 1000 gross
Therefore,
5 g = 5 g × 1000 gross / 3824 g
5 g = 1.3075 gross
Thus, 5 g is equivalent to 1.3075 gross.
Finally, we convert 1.3075 gross to dozen. This can be obtained as follow:
1 gross = 12 dozen
Therefore,
1.3075 gross = 1.3075 gross × 12 dozen / 1 gross
1.3075 gross = 15.69 dozen
Thus, 5 g of penny is equivalent to 15.69 dozen
volume of Ni = 25 nL = 25 x 10⁻⁹ L
mol Ni = 25 x 10⁻⁹ L x 1.25 mol/L = 3.125 x 10⁻⁸
mass = mol x Ar Ni
mass = 3.125 x 10⁻⁸ x 59 g/mol
mass = 1.84 x 10⁻⁶ g = 1.84 μg
Explanation:
A mixture in which there is uniform distribution of solute particles into the solvent is known as a homogeneous mixture.
For example, sugar dissolved in water is a homogeneous mixture.
On the other hand, a mixture in which there is uneven distribution of solute particles into the solvent is known as a heterogeneous mixture.
For example, sand present in water is a heterogeneous mixture.
Comment on given situations will be as follows.
(a) Air in a closed bottle - It is a homogeneous mixture because there will be even distribution of other gases that are present in air.
(b) Air over New York City - It is a heterogeneous mixture because there will be presence of some dust particles, fog or smoke into the air. Distribution of all these particles will be uneven. This will make air over New York City heterogeneous in nature.
The boiling point of water at 1 atm is 100 degrees celsius. However, when water is added with another substance the boiling point of it rises than when it is still a pure solvent. This called boiling point elevation, a colligative property. The equation for the boiling point elevation is expressed as the product of the ebullioscopic constant (0.52 degrees celsius / m) for water), the vant hoff factor and the concentration of solute (in terms of molality).
ΔT(CaCl2) = i x K x m = 3 x 0.52 x 0.25 = 0.39 °C
<span> ΔT(Sucrose) = 1 x 0.52 x 0.75 = 0.39 </span>°C<span>
</span><span> ΔT(Ethylene glycol) = 1 x 0.52 x 1 = 0.52 </span>°C<span>
</span><span> ΔT(CaCl2) = 3 x 0.52 x 0.50 = 0.78 </span>°C<span>
</span><span> ΔT(NaCl) = 2 x 0.52 x 0.25 = 0.26 </span>°C<span>
</span>
Thus, from the calculated values, we see that 0.75 mol sucrose dissolved on 1 kg water has the same boiling point with 0.25 mol CaCl2 dissolved in 1 kg water.
