Answer:
58.9mL
Explanation:
Given parameters:
Initial volume = 34.3mL = 0.0343dm³
Initial concentration = 1.72mM = 1.72 x 10⁻³moldm⁻³
Final concentration = 1.00mM = 1 x 10⁻³ moldm⁻³
Unknown:
Final volume =?
Solution:
Often times, the concentration of a standard solution may have to be diluted to a lower one by adding distilled water. To find the find the final volume, we must recognize that the number of moles of the substance in initial and final solutions are the same.
Therefore;
C₁V₁ = C₂V₂
where C and V are concentration and 1 and 2 are initial and final states.
now input the variables;
1.72 x 10⁻³ x 0.0343 = 1 x 10⁻³ x V₂
V₂ = 0.0589dm³ = 58.9mL
Answer:
Updrafts characterize a storm's early development, during which warm air rises to the level where condensation begins and precipitation starts to develop. In a mature storm, updrafts are present alongside downdrafts caused by cooling and by falling precipitation.
Hope it helps
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Answer: The correct answer is option (A).
Explanation:
Polar molecules are molecules in which formation of partial charges takes place due to which dipole moment gets created in a molecule. Molecules with polar bonds that s bond with partly ionic character. And water is of the example of polar molecule.
Electronegative oxygen atom in water molecule attracts the electron bond pair towards itself which generates partial negative charge on oxygen atom and partial positive charge on both hydrogen atoms.
Where as water has higher value surface tension due to strong intermolecular association of the water molecule due to presence of hydrogen bonding.And it is more denser is liquid state than in its solid state.
Hence,the correct answer is option (A).
Answer:
10 Litre
Explanation:
Given that ::
v1 = 25L ; n1 = 1.5 mole ; v2 =? ; n2 = (1.5-0.9) = 0.6 mole
Using the relation :
(n2 * v1) / n1 = (n2 * v2) / n2
v2 = (n2 * v1) / n1
v2 = (0.6 mole * 25 Litre) / 1.5 mole
v2 = 15 / 1.5 litre
v2 = 10 Litre
<u>We are given:</u>
The force applied on the poor hamster (F) = 12 N
Acceleration of the poor Hamster (a) = 8 m/s²
<u>Solving for the mass of the Poor Hamster:</u>
From newton's second equation of motion, we know that:
F = ma
<em>replacing the given values</em>
12 = 8 * m
m = 12/8 kg
m = 3/2 kg
The poor Hamster weighs 3/2 kg