Answer:
N = -3
H = +1
O = -2
Cl = +1
Explanation:
The balanced Equation of the reaction above is:
NH3 + 3HOCl ----> NCl3 + 3H2O
Where NH3 is ammonia;
HOCl is Oxochlorate (I) acid
NCl3 is nitrogen trichloride
H2O is water
Nitrogen in both Ammonia and Nitrogen trichloride has an oxidation state of -3. It shares three electrons with each of the three atoms of hydrogen and chlorine
Hydrogen in both Oxochlorate (I) acid and in water has an oxidation state of +1. Each atom shares an election with oxyen in both compounds.
Oxygen in both Oxochlorate (I) acid and water has an oxidation state of -2. Each atom shares two electrons with neighboring atoms; in HOCl, it shares with chlorine and hydrogen, while in H2O, it shares with two atoms of hydrogen.
Chlorine in both Oxochlorate (i) acid and Nitrogen trichloride has an oxidation state of +1. It shares an electron with each of it's neighboring bonded atoms.
Explanation:
The answer is the Calcium ion. It satisfies the conditions of the question.
Condition 1
The net ionic charge is one-tenth the nuclear charge.
In the Calcium ion, Ca²⁺. The nuclear charge in this ion is 20. The net ionic charge is 2.
2 / 20 = 1 / 10. So the net ionic charge is indeed one tenth of the nuclear charge.
Condition 2
The number of neutrons is four more than the number of electrons.
Mass Number of Ca²⁺ = 44
Atomic Number = 20
Neutrons = Mass Number - Atomic Number = 44 -20 = 24
Number of electrons = 20 - 2 = 18
Since Number of Neutrons = 22, Number of electrons = 18. This condition also holds.
As an ion and as an isotope = Ca²⁺, Ca - 42
Answer:
2.37x10⁻⁷ M⁻².s⁻¹
Explanation:
For a generic reversible reaction:
A + B ⇄ C + D
Kf is the constant of the formation of the products (C and D), Kr is the constant of the formation of the reactants (A and B), and Kc is the general equilibrium constant, which is:
Kc = Kf/Kr
2.76x10³ = 6.54x10⁻⁴/Kr
Kr = 6.54x10⁻⁴/2.76x10³
Kr = 2.37x10⁻⁷ M⁻².s⁻¹
Answer:
The answer is 42, 39 grams of LiCl
Explanation:
We calculate the weight of 1 mol of LiCl, from the atomic weights of each element obtained from the periodic table:
Weight 1 mol LiCl= Weight Li + Weight Cl =6,94 g+ 35, 45 g= 42, 39grams
The number of grams of Ag2SO4 that could be formed is 31.8 grams
<u><em> calculation</em></u>
Balanced equation is as below
2 AgNO3 (aq) + H2SO4(aq) → Ag2SO4 (s) +2 HNO3 (aq)
- Find the moles of each reactant by use of mole= mass/molar mass formula
that is moles of AgNO3= 34.7 g / 169.87 g/mol= 0.204 moles
moles of H2SO4 = 28.6 g/98 g/mol =0.292 moles
- use the mole ratio to determine the moles of Ag2SO4
that is;
- the mole ratio of AgNo3 : Ag2SO4 is 2:1 therefore the moles of Ag2SO4= 0.204 x1/2=0.102 moles
- The moles ratio of H2SO4 : Ag2SO4 is 1:1 therefore the moles of Ag2SO4 = 0.292 moles
- AgNO3 is the limiting reagent therefore the moles of Ag2SO4 = 0.102 moles
<h3> finally find the mass of Ag2SO4 by use of mass=mole x molar mass formula</h3>
that is 0.102 moles x 311.8 g/mol= 31.8 grams
