Answer: the line Spectra of hydrogen lies between the ultra-violet, visible light and infra-red of the electro magnetic spectrum
Explanation:
Electromagnetic radiation spans an wide range of wavelengths and frequencies. This range is called the electromagnetic spectrum. The electromagnetic spectrum is generally divided into seven regions, in order of decreasing wavelength and increasing energy and frequency. The 7 regions includes; radio waves, microwaves, infrared (IR), visible light, ultraviolet (UV), X-rays and gamma rays.
lower-energy radiation, such as radio waves, is expressed as frequency while microwaves, infrared, visible and UV light are usually expressed as wavelength and finally, higher-energy radiation such as X-rays and gamma rays, is expressed in terms of energy per photon.
Therefore, hydrogen lies between the ultra-violet, visible light and infra-red region of the electro magnetic spectrum.
Quantitative is a description in numbers , and qualitative is a description with words
Answer : q = 6020 J, w = -6020 J, Δe = 0
Solution : Given,
Molar heat of fusion of ice = 6020 J/mole
Number of moles = 1 mole
Pressure = 1 atm
Molar heat of fusion : It is defined as the amount of energy required to melt 1 mole of a substance at its melting point. There is no temperature change.
The relation between heat and molar heat of fusion is,
(in terms of mass)
or, 
(in terms of moles)
Now we have to calculate the value of q.
When temperature is constant then the system behaves isothermally and Δe is a temperature dependent variable.
So, the value of 
Now we have to calculate the value of w.
Formula used : 
where, q is heat required, w is work done and 
is internal energy.
Now put all the given values in above formula, we get
w = -6020 J
Therefore, q = 6020 J, w = -6020 J, Δe = 0
There are 19.5 g Na in 71.4 g NaHCO₃
Calculate the <em>molecular mass of NaHCO₃</em>.
1 Na = 1 × 22.99 u = 22.99 u
1 H = 1 × 1.008 u = 1.008 u
1 C = 1 × 12.01 u = 12.01 u
3 O = 3 × 16.00 u = <u>48.00 u
</u>
TOTAL = 84.008 u
So, there are 22.99 g of Na in 84.008 g NaHCO₃.
∴ Mass of Na = 71.4 g NaHCO₃ × (22.99 g Na/84.008 g NaHCO₃) = 19.5 g Na
