Explanation:
Moles of compound =
We have ;
Volume of solution = 600 mL = 0.600 L ( 1 mL = 0.001 L)
Moles of NaOH = n
Molarity of the solution = 3 M
n = 3 M × 0.600 L = 1.800 mol
Mass of 1.800 mole sof NaOH :
1.800 mol × 40 g/mol = 72.0 g
Preparation:
Weight 72.0 grams of sodium hydroxide and add it to the 500 mL of volumetric flask along with some water. Dissolve the all the solute by adding small proportion of water. After the solution becomes clear make the water upto the mark of 500 ml.
Transfer the solution to a bigger beaker and 100 mL of water more to it.
(i) We start by calculating the mass of sugar in the solution:
mass of sugar = concentration × solution mass
mass of sugar = 2.5/100 × 500 = 12.5 g
Then now we can calculate the amount of water:
solution mass = mass of sugar + mass of water
mass of water = solution mass - mass of sugar
mass of water = 500 - 12.5 = 487.5 g
(ii) We use the following reasoning:
If 500 g solution contains 12.5 g sugar
Then X g solution contains 75 g sugar
X=(500×75)/12.5 = 3000 g solution
Now to get the amount of solution in liters we use density (we assume that is equal to 1):
Density = mass / volume
Volume = mass / density
Volume = 3000 / 1 = 3000 liters of sugar solution
Answer:
E = 3.77×10⁻¹⁹ J
Explanation:
Given data:
Wavelength of absorption line = 527 nm (527×10⁻⁹m)
Energy of absorption line = ?
Solution:
Formula:
E = hc/λ
h = planck's constant = 6.63×10⁻³⁴ Js
c = speed of wave = 3×10⁸ m/s
by putting values,
E = 6.63×10⁻³⁴ Js × 3×10⁸ m/s / 527×10⁻⁹m
E = 19.89×10⁻²⁶ Jm /527×10⁻⁹m
E = 0.0377×10⁻¹⁷ J
E = 3.77×10⁻¹⁹ J
Answer:-
The reaction of 2-bromopropane reacts with sodium iodide in acetone is an example of Sn2 reaction.
The I - attacks from backside to give the transition state for both.
If we compare the transition state for cyclobromopropane 2-bromopropane then we see in case of cyclobromopropane transition state, one of the H is very close to the incoming I -.
This results in steric strain and less stability of the transition state. Hence 2-bromopropane reacts with sodium iodide in acetone over 104 times faster than bromocyclopropane.
3 Chlorine ions are required to bond with one aluminum ion.
In ionic bonds, metals atoms loses all its outermost shell electrons to form a cation. While, non metal atoms gains however many electrons in order to make its outermost electron shell be 8 (or 2 if there's only one shell).
Therefore, form the periodic table, we can see that aluminum has a atomic number of 13, which makes its electron arrangement be 2,8,3. So, in order to form a aluminum ion, an Al atom must lose 3 electrons. On the other hand, Chlorine has a atomic number of 17, which means it has the electron configuration of 2,8,7. It has to gain only 1 electron to have 8 outermost shell electron.
Thereofre, 3 Chlorine atom are required to gain all 3 electrons given out by just 1 aluminum ion.
