Answer is: an oxybromate compound is KBrO₄ (x = 4).
ω(Br) = 43.66% ÷ 100%.
ω(Br) = 0.4366; mass percentage of bromine.
If we take 100 grams of compound:
m(Br) = ω(Br) · 100 g.
m(Br) = 0.4366 · 100 g.
m(Br) = 43.66 g; mass of bromine.
n(Br) = m(Br) ÷ M(Br).
n(Br) = 43.66 g ÷ 79.9 g/mol,
n(Br) = 0.55 mol; amoun of bromine.
From chemical formula (KBrOₓ), amount of potassium is equal to amount of bromine: n(Br) = n(K).
m(K) = 0.55 mol · 39.1 g/mol.
m(K) = 21.365 g; mass of potassium in the compound.
m(O) = 100 g - 21.365 g - 43.66 g.
m(O) =34.97 g; mass of oxygen.
n(O) = 34.97 g ÷ 16 g/mol.
n(O) = 2.185 mol.
n(K) : n(Br) : n(O) = 0.55 mol : 0.55 mol : 2.185 mol /÷ 0.55 mol.
n(K) : n(Br) : n(O) = 1 : 1 : 4.
<h3>
Answer:</h3>
3CaCl₂ + 2Na₃PO₄→ Ca₃(PO₄)₂ + 6NaCl
<h3>
Explanation:</h3>
We are given the Equation;
CaCl₂ + Na₃PO₄→ Ca₃(PO₄)₂ + NaCl
Assuming the question requires us to balance the equation;
- A balanced chemical equation is one that has equal number of atoms of each element on both sides of the equation.
- Balancing chemical equations ensures that they obey the law of conservation of mass in chemical equations.
- According to the law of conservation of mass in chemical equation, the mass of the reactants should always be equal to the mass of the products.
- Balancing chemical equations involves putting appropriate coefficients on the reactants and products.
In this case;
- To balance the equation we are going to put the coefficients 3, 2, 1, and 6.
- Therefore; the balanced equation will be;
3CaCl₂ + 2Na₃PO₄→ Ca₃(PO₄)₂ + 6NaCl
C.) Ionic bond is formed between that...
Answer:
Neutral solution is formed.
Explanation:
When the hydrochloric acid and sodium hydroxide which is a strong base are combined together, it produces sodium chloride which is a salt and water. This solution is known as Neutral solution because the solution do not have the characteristics or properties of either an acid or a base. If the concentration of one of the reactant is higher as compared to another reactant so the product has the characteristics of that reactant.
Answer:
Kc = 1.54e - 31 / 2.61e - 24
Explanation:
1 ) 
; Kc = 1.54e - 31
2) 
; Kc = 2.16e - 24
upon reversing ( 2 ) equation
Kc = 1/2.16e - 24
now adding 1 and reversed equation (2)
we get ,
Kc = 1.54e-31 × 1/2.61e - 24
equilibrium constant of equation (3) is -
Kc = 1.54e - 31 / 2.61e - 24
