Answer:
Specific heat of water = 33.89 KJ
Explanation:
Given:
mass of water = 81 gram
Initial temperature = 0°C
Final temperature = 100°C
Specific heat of water = 4.184
Find:
Required heat Q
Computation:
Q = Mass x Specific heat of water x (Final temperature - Initial temperature)
Q = (81)(4.184)(100-0)
Q = 33,890.4
Specific heat of water = 33.89 KJ
Homemade milkshake(banana&strawberry)
Answer:
Frequency = 6.16 ×10¹⁴ Hz
λ = 4.87×10² nm
Explanation:
In case of hydrogen atom energy associated with nth state is,
En = -13.6/n²
For n = 2
E₂ = -13.6 / 2²
E₂ = -13.6/4
E₂ = -3.4 ev
Kinetic energy of electron = -E₂ = 3.4 ev
For n = 4
E₄ = -13.6 / 4²
E₄ = -13.6/16
E₄ = -0.85 ev
Kinetic energy of electron = -E₄ = 0.85 ev
Wavelength of radiation emitted:
E = hc/λ = E₄ - E₂
hc/λ = E₄ - E₂
by putting values,
6.63×10⁻³⁴Js × 3×10⁸m/s / λ = -0.85ev - (-3.4ev )
6.63×10⁻³⁴ Js× 3×10⁸m/s / λ = 2.55 ev
λ = 6.63×10⁻³⁴ Js× 3×10⁸m/s /2.55ev
λ = 6.63×10⁻³⁴ Js× 3×10⁸m/s /2.55× 1.6×10⁻¹⁹ J
λ = 19.89 ×10⁻²⁶ Jm / 2.55× 1.6×10⁻¹⁹ J
λ = 19.89 ×10⁻²⁶ Jm / 4.08×10⁻¹⁹ J
λ = 4.87×10⁻⁷ m
m to nm:
4.87×10⁻⁷ m ×10⁹nm/1 m
4.87×10² nm
Frequency:
Frequency = speed of electron / wavelength
by putting values,
Frequency = 3×10⁸m/s /4.87×10⁻⁷ m
Frequency = 6.16 ×10¹⁴ s⁻¹
s⁻¹ = Hz
Frequency = 6.16 ×10¹⁴ Hz
The pH of a 0.260 M solution of ascorbic acid is 0.585. Details about pH can be found below.
<h3>How to calculate pH?</h3>
The pH of a solution can be calculated using the following expression:
pH = - log {H+}
According to this question, ascorbic acid is a diprotic acid and posseses a concentration of 0.260M. The pH can be calculated as follows;
pH = - log {0.260}
pH = 0.585
Therefore, the pH of a 0.260 M solution of ascorbic acid is 0.585.
Learn more about pH at: brainly.com/question/15289741
#SPJ1
I think it’s to long to fit in a period??
