Answer:
1. Mg (s) + 2Na+(aq) → 2Na(s) + Mg²⁺(aq)
2. 2K(s) + Cd²⁺(aq) → 2K⁺(aq) + Cd(s)
Explanation:
The net ionic equation of a reaction express only the chemical species that are involved in the reaction:
1. Mg (s) + Na2CrO4 (aq) → 2Na + MgCrO4(aq)
The ionic equation:
Mg (s) + 2Na+(aq) + CrO4²⁻ (aq) → 2Na + Mg²⁺ + CrO4²⁻(aq)
Subtracting the ions that don't change:
<h3>Mg (s) + 2Na+(aq) → 2Na + Mg²⁺</h3>
2. 2K(s) + Cd(NO3)2(aq) → 2KNO3(aq) + Cd(s)
The ionic equation:
2K(s) + Cd²⁺(aq) + 2NO3⁻(aq) → 2K⁺(aq) + 2NO3⁻(aq) + Cd(s)
Subtracting the ions that don't change:
<h3>2K(s) + Cd²⁺(aq) → 2K⁺(aq) + Cd(s)</h3>
<u>Answer:</u> The final temperature of water is 32.3°C
<u>Explanation:</u>
When two solutions are mixed, the amount of heat released by solution 1 (liquid water) will be equal to the amount of heat absorbed by solution 2 (liquid water)
The equation used to calculate heat released or absorbed follows:
![m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]](https://tex.z-dn.net/?f=m_1%5Ctimes%20c%5Ctimes%20%28T_%7Bfinal%7D-T_1%29%3D-%5Bm_2%5Ctimes%20c%5Ctimes%20%28T_%7Bfinal%7D-T_2%29%5D)
......(1)
where,
q = heat absorbed or released
= mass of solution 1 (liquid water) = 50.0 g
= mass of solution 2 (liquid water) = 29.0 g
= final temperature = ?
= initial temperature of solution 1 = 25°C = [273 + 25] = 298 K
= initial temperature of solution 2 = 45°C = [273 + 45] = 318 K
c = specific heat of water= 4.18 J/g.K
Putting values in equation 1, we get:
![50.0\times 4.18\times (T_{final}-298)=-[29.0\times 4.18\times (T_{final}-318)]\\\\T_{final}=305.3K](https://tex.z-dn.net/?f=50.0%5Ctimes%204.18%5Ctimes%20%28T_%7Bfinal%7D-298%29%3D-%5B29.0%5Ctimes%204.18%5Ctimes%20%28T_%7Bfinal%7D-318%29%5D%5C%5C%5C%5CT_%7Bfinal%7D%3D305.3K)
Converting this into degree Celsius, we use the conversion factor:
Hence, the final temperature of water is 32.3°C
Formula : BaI₂. <span>barium iodide</span>
Answer:
11.31 g.
Explanation:
Molarity is defined as the no. of moles of a solute per 1.0 L of the solution.
M = (no. of moles of solute)/(V of the solution (L)).
<em>∴ M = (mass/molar mass)of NaCl/(V of the solution (L)).</em>
<em></em>
<em>∴ mass of NaCl remained after evaporation of water = (M)(V of the solution (L))(molar mass)</em> = (0.45 M)(0.43 L)(58.44 g/mol) = <em>11.31 g.</em>
