1. Ionic Bond
2.It is a chemical reaction because the reactants react to create products and it is endothermic because it takes in heat from the sunlight which helps it to photosynthesise
Hope that helps ☻
Total number of students (male and female) in the college=2760. Total number of students including both male and female represents 100% (percentage) of students which is equal to 2760 number.
Number of male students is 65% (percentage) which is equal to
= 1794. So, total number of female students are (100-65)%=35% (percentage) which is equal to
=966 numbers.
Question in incomplete, complete question is:
Technetium (Tc; Z = 43) is a synthetic element used as a radioactive tracer in medical studies. A Tc atom emits a beta particle (electron) with a kinetic energy (Ek) of
. What is the de Broglie wavelength of this electron (Ek = ½mv²)?
Answer:
is the de Broglie wavelength of this electron.
Explanation:
To calculate the wavelength of a particle, we use the equation given by De-Broglie's wavelength, which is:

where,
= De-Broglie's wavelength = ?
h = Planck's constant = 
m = mass of beta particle = 
= kinetic energy of the particle = 
Putting values in above equation, we get:


is the de Broglie wavelength of this electron.
Answer:
(NH4)2CO3 is the formula !