<h3>1</h3>
Species shown in bold are precipitates.
- Ca(NO₃)₂ + 2 KOH → Ca(OH)₂ + 2 KNO₃
- Ca(NO₃)₂ + Na₂C₂O₄ → CaC₂O₄ + 2 NaNO₃
- Cu(NO₃)₂ + 2 KI → CuI₂ + 2 KI
- Cu(NO₃)₂ + 2 KOH → Cu(OH)₂ + 2 KNO₃
- Cu(NO₃)₂ + Na₂C₂O₄ → CuC₂O₄ + 2 NaNO₃
- Ni(NO₃)₂ + 2 KOH → Ni(OH)₂ + 2 KNO₃
- Ni(NO₃)₂ + Na₂C₂O₄ → NiC₂O₄ + 2 NaNO₃
- Zn(NO₃)₂ + 2 KOH → Zn(OH)₂ + 2 KNO₃
- Zn(NO₃)₂ + Na₂C₂O₄ → ZnC₂O₄ + 2 NaNO₃
<h3>2</h3>
A double replacement reaction takes place only if it reduces in the concentration of ions in the solution. For example, the reaction between Ca(NO₃)₂ and KOH produces Ca(OH)₂. Ca(OH)₂ barely dissolves. The reaction has removed Ca²⁺ and OH⁻ ions from the solution.
Some of the reactions lead to neither precipitates nor gases. They will not take place since they are not energetically favored.
<h3>3</h3>
Compare the first and last row:
Both Ca(NO₃)₂ and Zn(NO₃)₂ react with KOH. However, between the two precipitates formed, Ca(OH)₂ is more soluble than Zn(OH)₂.
As a result, add the same amount of KOH to two Ca(NO₃)₂ and Zn(NO₃)₂ of equal concentration. The solution that end up with more precipitate shall belong to Zn(NO₃)₂.
<h3>4</h3>
Compare the second and third row:
Cu(NO₃)₂ reacts with KI, but Ni(NO₃)₂ does not. Thus, add equal amount of KI to the two unknowns. The solution that forms precipitate shall belong to Cu(NO₃)₂.
Answer:
V₂ = 5.97 L
Explanation:
Given data:
Initial temperature = 9°C (9+273 = 282 K)
Initial volume of gas = 6.17 L
Final volume of gas = ?
Final temperature = standard = 273 K
Solution:
Formula:
The Charles Law will be apply to solve the given problem.
According to this law, 'the volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure'
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 6.17 L × 273K / 282 k
V₂ = 1684.41 L.K / 282 K
V₂ = 5.97 L
The mass of lime that can be produced from 4.510 Kg of limestone is calculated as below
calculate the moles of CaCO3 used
that is moles =mass/molar mass
convert Kg to g = 4.510 x1000 =4510g
= 4510 / 100 =45.10 moles
CaCO3 = CaO +O2
by use of mole ratio between CaCO3 to CaO (1:1) the moles of CaO is also= 45.10 moles
mass of CaO = moles x molar mass
45.10 x56 = 2525.6 g of CaO
4Al(s) + 3O2(g) --> 2Al2O3(s) This is the balanced.
From the equation:
4 moles of Al required 3 moles of O2 to produce 2 moles of Al2O3
3 moles of O2 reacted with 4 moles of Al to produce 2 moles of Al2O3
1 mole of O2 reacted with 4/3 moles of Al to produce 2/3 moles of Al2O3 (Divide by 3)
4.5 moles of O2 reacted with (4/3 *4.5) moles of Al to produce (2/3*4.5) moles of Al2O3
4.5 moles of O2 reacted with 6moles of Al to produce 3moles of Al2O3
(3) is the answer. 6 mol of Al.
