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Understanding how much of a product is produced in a reaction is referred to as Stoichiometrically understanding the reaction.
Stoichiometry is the calculation of the quantitative relationships between reactants and products in a chemical reaction. The first to talk about stoichiometry was Jeremias Benjamin Ritcher, who said that "Stoichiometry is the science that measures the quantitative proportions or mass ratios of chemical elements that are involved in a chemical reaction".
To calculate how much of a product is produced in a reaction, Stoichiometry is used, applying the law of conservation of mass. That means that the amount of product can be calculated from the amounts of reactants if they are known.
Answer:
k = -0.09165 years^(-1)
Explanation:
The exponential decay model of a radioactive isotope is generally given as;
A(t) = A_o(e^(kt))
Where;
A_o is quantity of isotope before decay, k is decay constant and A(t) is quantity after t years
We are given;
A_o = 5 kg
A(10) = 2kg
t = 10 years
Thus;
A(10) = 2 = 5(e^(10k))
Thus;
2 = 5(e^(10k))
2/5 = (e^(10k))
0.4 = (e^(10k))
In 0.4 = 10k
-0.9164 = 10k
k = -0.9164/10
k = -0.09165 years^(-1)
Valence électrons = are electrons in the outermost shell that is responsible for the chemical reactions of the atoms
Explanation:
The Bohr hypothesis shows that an atom has an circular orbit of electrons outside the nucleus in a revolving orbit. In Bohr model for Nitrogen there are two orbits s and p. The s has two electron and p has five. Whereas in Bohr neon model there are also orbital s, and p.. The s orbital has 2 electron and the p orbital has 8 electron in it.
Answer:
HNO₃
Explanation:
Data given
Nitrogen = 9.8 g
Hydrogen = 0.70 g
Oxygen = 33.6 g
Empirical formula = ?
Solution:
Convert the masses to moles
For Nitrogen
Molar mass of N = 14 g/mol
no. of mole = mass in g / molar mass
Put value in above formula
no. of mole = 9.8 g/ 14 g/mol
no. of mole = 0.7
mole of N = 0.7 mol
For Hydrogen
Molar mass of H = 1 g/mol
no. of mole = mass in g / molar mass
Put value in above formula
no. of mole = 0.70 g/ 1 g/mol
no. of mole = 0.7
mole of H = 0.7 mol
For Oxygen
Molar mass of O = 16 g/mol
no. of mole = mass in g / molar mass
Put value in above formula
no. of mole = 33.6 g / 16 g/mol
no. of mole = 2.1
mole of O = 2.1 mol
Now we have values in moles as below
N = 0.7
H = 0.7
O = 2.1
Divide the all values on the smallest values to get whole number ratio
N = 0.7 / 0.7 = 1
H = 0.7 / 0.7 = 1
O = 2.1 / 0.7 = 3
So all have following values
N = 1
H = 1
O = 3
So the empirical formula will be HNO₃ i.e. all three atoms in simplest small ratio.
