Answer:
electron
Explanation:
The electron is subatomic particle that revolve around outside the nucleus and has negligible mass. It has a negative charge.
Symbol = e⁻
Mass = 9.10938356×10⁻³¹ Kg
Mass in amu = 1/1838 = 5.4 × 10⁻⁴amu
It was discovered by j. j. Thomson in 1897 during the study of cathode ray properties.
While neutron and proton are present inside the nucleus. Proton has positive charge while neutron is electrically neutral. Proton is discovered by Rutherford while neutron is discovered by James Chadwick in 1932.
Symbol of proton= P⁺
Symbol of neutron= n⁰
Mass of proton=1.672623×10⁻²⁷ Kg
Mass of neutron=1.674929×10⁻²⁷Kg
An atom consist of electron, protons and neutrons. Protons and neutrons are present with in nucleus while the electrons are present out side the nucleus.
All these three subatomic particles construct an atom
The concentration of the drug stock solution is 1.5*10^-9 M i.e. 1.5 * 10^-9 moles of the drug per Liter of the solution
Therefore, the number of moles present in 1 ml i.e. 1*10^-3 L of the solution would be = 1 *10^-3 L * 1.5 * 10^-9 moles/1 L = 1.5 * 10^-12 moles
1 mole of the drug will contain 6.023*10^23 drug molecules
Therefore, 1.5*10^-12 moles of the drug will correspond to :
1.5 * 10^-12 moles * 6.023*10^23 molecules/1 mole = 9.035 * 10^11 molecules
The number of cancer cells = 2.0 * 10^5
Hence the ratio = drug molecules/cancer cells
= 9.035 *10^11/2.0 *10^5
= 4.5 * 10^6
In nature, boron is monoatomic. Therefore, its formula is B.
On the other hand, fluorine is diatomic. Therefore, its formula is F2
Now, the basic unbalanced equation is:
B + F2 .........> BF3
Now, we need to balance this equation. As you can see, we have two fluorine moles entering the reaction and 3 formed in the products.
Balancing the equation, we will reach the following balanced reaction:
2B + 3F2 .......> 2BF3
The element used in advertising signs is neon.
Answer:
The mass percentage of calcium carbonated reacted is 2.5%.
Explanation:
The reaction is:
Thus the Kp of the equilibrium will be:
Kp = partial pressure of carbon dioxide [as the other are solid]
Moles of calcium carbonate initially present = 
Let us apply ICE table to the equilibrium given:
Initial 0.2 0 0
Change -x +x +x
Equilibrium 0.2-x x x
Kp = partial pressure of carbon dioxide
Kp = Kc(RT)ⁿ
where n = difference in the number of moles of gaseous products and reactants
for given reaction n = 1
R = gas constant = 8.314 J /mol K
T = temperature = 800 ⁰C = 1073 K
Putting values
Kc =
Kc = ![\frac{[CO_{2}][CaO]}{[CaCO_{3}]}= \frac{x^{2} }{(0.2-x)}=1.3X10^{-4}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCO_%7B2%7D%5D%5BCaO%5D%7D%7B%5BCaCO_%7B3%7D%5D%7D%3D%20%5Cfrac%7Bx%5E%7B2%7D%20%7D%7B%280.2-x%29%7D%3D1.3X10%5E%7B-4%7D)
On calculating
x = 0.005
where x = the moles of calcium carbonate dissociated or reacted.
Percentage of the moles or mass reacted = 
%
