Answer:
v = 7.3 × 10⁶ m/s
Explanation:
Given data:
Velocity of electron = ?
Wavelength = 100 pm
Solution:
Formula:
λ = h/mv
λ = wavelength
h = planck's constant
m = mass
v = velocity
Now we will put the values in formula.
100 ×10⁻¹² m = 6.63 × 10⁻³⁴ j.s / 9.109 × 10⁻³¹ kg × v
v = 6.63 × 10⁻³⁴ kg.m²/s / 9.109 × 10⁻³¹ kg ×100 ×10⁻¹² m
v = 6.63 × 10⁻³⁴ m/s /910.9 × 10⁻⁴³
v = 0.0073 × 10⁹ m/s
v = 7.3 × 10⁶ m/s
Answer:
We can do the nitration of benzene by treating the benzene with a mixture of nitric acid and sulphuric acid by not extending the temperature of 50°C
Explanation:
Nitration of benzene takes place by treating the benzene with a mixture of nitric acid and sulphuric acid at low temperatures such as the temperatures below 50°C
The nitration of benzene takes place through electrophilic substitution reaction
In this reaction the electrophile is nitronium ion (NO2+) which performs an electrophilic substitution reaction on the benzene ring and during the reaction an intermediate will also be formed in which there will be positive charge distributed in the benzene
These electrophile is generated when nitric acid is treated with sulphuric acid
As nitric acid is a strong oxidising agent, here in this case the oxidation state of nitrogen will change from +5 to +3
The reactions regarding the nitration of benzene is present in the file attached
Answer: 0.529 atm
Explanation:
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,
where,
= initial pressure of gas = 0.998 atm
= final pressure of gas = ?
= initial volume of gas = 2 L
= final volume of gas = 3.5 L
= initial temperature of gas =
= final temperature of gas =
Now put all the given values in the above equation, we get:
Thus the pressure if it is brought to a higher altitude where it now occupies 3.5 L and is at 12.0 °C is 0.529 atm
True would be your answer
Answer:
A. 8.8 g.
B. 5.3 g.
C. 449 g.
Explanation:
A. Determination of the mass of CO2.
Mole of CO2 = 0.2 mole
Molar mass of CO2 = 12 + (2×16)
= 12 + 32
= 44 g/mol
Mass of CO2 =?
Mole = mass /Molar mass
0.2 = mass of CO2 /44
Cross multiply
Mass of CO2 = 0.2 × 44
Mass of CO2 = 8.8 g
B. Determination of the mass of Na2CO3.
Mole of Na2CO3 = 0.05 mole
Molar mass of Na2CO3 = (2×23) + 12 + (3×16)
= 46 + 12 + 48
= 106 g/mol
Mass of Na2CO3 =?
Mole = mass /Molar mass
0.05 = mass of Na2CO3 /106
Cross multiply
Mass of Na2CO3 = 0.05 × 106
Mass of Na2CO3 = 5.3 g
C. Determination of the mass of Fe(OH)2.
Mole of Fe(OH)2 = 5 moles
Molar mass of Fe(OH)2 = 55.8 + 2(16 + 1)
= 55.8 + 2(17)
= 55.8 + 34
= 89.8 g/mol
Mass of Fe(OH)2 =?
Mole = mass /Molar mass
5 = mass of Fe(OH)2/89.8
Cross multiply
Mass of Fe(OH)2 = 5 × 89.9
Mass of Fe(OH)2 = 449 g.