Answer:
The number of positive charges in nucleus of an atoms are equal to the atomic number and also positive charges are equal to the negative charges which are electrons in neutral atom.
Explanation:
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.
Electron:
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
It was discovered by j. j. Thomson in 1897 during the study of cathode ray properties.
Neutron and proton:
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=1.674929×10⁻²⁷ Kg
All these three subatomic particles construct an atom. A neutral atom have equal number of proton and electron. In other words we can say that negative and positive charges are equal in magnitude and cancel the each other. For example if neutral atom has 6 protons than it must have 6 electrons. The sum of neutrons and protons is the mass number of an atom while the number of protons are number of electrons is the atomic number of an atom.
For example
The carbon have six protons and six neutrons so its atomic mass is 12 amu and atomic number is six.
Answer:
Amount of excess Carbon (ii) oxide left over = 23.75 g
Explanation:
Equation of the reaction: Fe₂O₃ + 3CO ----> 2Fe + 3CO₂
Molar mass of Fe₂O₃ = 160 g/mol;
Molar mass of Carbon (ii) oxide = 28 g/mol
From the equation of reaction, 1 mole of Fe₂O₃ reacts with 3 moles of carbon (ii) oxide; i.e. 160 g of iron (iii) oxide reacts with 84 g (3 * 28 g) of carbon (ii) oxide
450 g of Fe₂O₃ will react with 450 * 84/180) g of carbon (ii) oxide = 236..25 g of carbon (ii) oxide
Therefore the excess reactant is carbon (ii) oxide.
Amount of excess Carbon (ii) oxide left over = 260 - 236.25
Amount of excess Carbon (ii) oxide left over = 23.75 g
Answer:
Explanation:
Density is
mass / volume = d
Mass:
840g
Volume:
7 cm x 4 cm x 10 cm = 280 cm^3
840g / 280 cm^3 = 3 g/cm^3
The highest point is the crest
The number of moles in monochloro pentaborane (9) 
is 0.37056.
We know that ,
Number of moles = Given mass/ Molar mass
Given mass of monochloro pentaborane (9) is 20.1 g
The molar mass of monochloro pentaborane (9) is
= 
= 
= 54.241
Thus number of moles = 20.1/54.241
= 0.37056
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