Answer:
The coefficient of Ca(OH)2 is 1
Explanation:
Step 1: unbalanced equation
Ca(OH)2 + HNO3 → Ca(NO3)2 + H2O
Step 2: Balancing the equation
On the right side we have 2x N (in Ca(NO3)2 ) and 1x N on the left side (in HNO3). To balance the amount of N on both sides, we have to multiply HNO3 by 2.
Ca(OH)2 + 2HNO3 → Ca(NO3)2 + H2O
On the left side we have 4x H (2xH in Ca(OH)2 and 2x H in HNO3), on the right side we have 2x H (in H2O). To balance the amount of H on both sides, we have to multiply H2O on the right side, by 2.
Now the equationis balanced.
Ca(OH)2 + 2HNO3 = Ca(NO3)2 + 2H2O
The coefficient of Ca(OH)2 is 1
Answer:
2K (s) + Cl₂ (g) ⇒ 2KCl (s)
Explanation:
Potassium and chlorine gas combine to form potassium chloride which is an ionic compound. The reaction is a type of combination reaction in which chlorine is being added to the metal, potassium.
Potassium reacts violently with the chlorine which is yellowish green in color to produce white solid of potassium chloride.
The balanced reaction is shown below as:
2K (s) + Cl₂ (g) ⇒ 2KCl (s)
Answer : The molar mass of the solute would be low.
Explanation :
Formula used for depression in freezing point is:

where,
= change in freezing point
= freezing point of solution
= freezing point of water
i = Van't Hoff factor
= freezing point constant
m = molality
= mass of solute
= mass of solvent
= molar mass of solute
From the formula we conclude that, when the freezing point of the solution read incorrectly that is freezing point of the solution is lower than the true freezing point then this means that change in freezing point would be high and the molar mass of the solute would be low.
Hence, the molar mass of the solute would be low.
Answer:
think I did this before and its V
Answer:
[Zn²⁺] = 4.78x10⁻¹⁰M
Explanation:
Based on the reaction:
ZnBr₂(aq) + K₂CO₃(aq) → ZnCO₃(s) + 2KBr(aq)
The zinc added produce the insoluble ZnCO₃ with Ksp = 1.46x10⁻¹⁰:
1.46x10⁻¹⁰ = [Zn²⁺] [CO₃²⁻]
We can find the moles of ZnBr₂ added = Moles of Zn²⁺ and moles of K₂CO₃ = Moles of CO₃²⁻ to find the moles of CO₃²⁻ that remains in solution, thus:
<em>Moles ZnB₂ (Molar mass: 225.2g/mol) = Moles Zn²⁺:</em>
6.63g ZnBr₂ * (1mol / 225.2g) = 0.02944moles Zn²⁺
<em>Moles K₂CO₃ = Moles CO₃²⁻:</em>
0.100L * (0.60mol/L) = 0.060 moles CO₃²⁻
Moles CO₃²⁻ in excess: 0.0600moles CO₃²⁻ - 0.02944moles =
0.03056moles CO₃²⁻ / 0.100L = 0.3056M = [CO₃²⁻]
Replacing in Ksp expression:
1.46x10⁻¹⁰ = [Zn²⁺] [0.3056M]
<h3>[Zn²⁺] = 4.78x10⁻¹⁰M</h3>