The product and balanced net ionic equations for the following reactions are SnCl₂ + 2KMnO₄ ⇒ 2 KCl + Sn(MnO₄)₂.
Ionic equations are those equations that happened in an aqueous solution. The chemical equation is expressed in an electrolyte solution is expressed and dissociates ions.
In these reactions, each element or ion is dissociated into differently charged ions in a solution. Each of the ions is shown with different charges.
SnCl₂ + 2KMnO₄ ⇒ 2 KCl + Sn(MnO₄)₂. In the reaction, the tin chloride, and potassium magnesium oxide. It dissociates into charged ions, like potassium chloride and tin magnesium oxide. The chlorine will acquire a negative charge and magnesium oxide get a positive charge.
Thus, the net ionic equation is SnCl₂ + 2KMnO₄ ⇒ 2 KCl + Sn(MnO₄)₂.
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Answer:
The answer to your question is:
Explanation:
Reaction
SnCl₂ + 2KMnO₄ ⇒ 2 KCl + Sn(MnO₄)₂
1 ---- Sn ---- 1
2 ---- K ----- 2
2 ---- Mn ---- 2
8 ---- O ---- 8
2 ---- Cl ---- 2
Answer:
NONE OF THE ABOVE
Explanation:
None of the above are examples of an oxidation - reduction or a redox reaction . This is because there is no change in the oxidation state of any of the elements in the reaction when the reaction happens .
⇒
Answer:
0.2 moles of CO₂ are produced
Explanation:
Given data:
Moles of CO₂ produced = ?
Moles of Na₂CO₃ react = 0.2 mol
Solution:
Chemical equation:
Na₂CO₃ + 2HCl → 2NaCl + CO₂ + H₂O
Now we will compare the moles of CO₂ with Na₂CO₃ .
Na₂CO₃ : CO₂
1 : 1
0.2 : 0.2
Thus, 0.2 moles of CO₂ are produced.
a.carbon?
Answer : The concentration of ion, pH and pOH of solution is, , 4.98 and 9.02 respectively.
Explanation : Given,
Concentration of ion =
pH : It is defined as the negative logarithm of hydrogen ion or hydronium ion concentration.
The expression used for pH is:
First we have to calculate the pH.
The pOH of the solution is, 9.02
Now we have to calculate the pH.
The pH of the solution is, 4.98
Now we have to calculate the concentration.
The concentration is,
Answer:
pOH = 9.022, [H⁺] = 1.5×10⁻⁵ M, pH = 4.978
Explanation:
Given: [OH⁻] = 9.5 × 10⁻¹⁰ M, T= 25°C
As, pOH = - log [OH⁻]
⇒ pOH = - log (9.5 x 10⁻¹⁰) = 9.022
The self-ionisation constant of water is given by
Kw = [H⁺] [OH⁻] and pKw = pH + pOH
Since, at room temperature (25°C): Kw = 1.0 × 10⁻¹⁴ and pKw = 14.
Therefore, Kw = [H⁺] [OH⁻] = 1.0 × 10⁻¹⁴
⇒ [H⁺] = (1.0 × 10⁻¹⁴) ÷ [OH⁻] = (1.0 ×10⁻¹⁴) ÷ [9.5 × 10⁻¹⁰] = 0.105 ×10⁻⁴ = 1.5×10⁻⁵ M
also,
pH + pOH = pKw = 14
⇒ pH = 14 - pOH = 14 - 9.022 = 4.978
Answer:
water and salt
Explanation:
In the FLVS salt will be your answer but technaically when a acid reacts with a base it can create salt and water