Answer:
We need 1.1 grams of Mg
Explanation:
Step 1: Data given
Volume of water = 78 mL
Initial temperature = 29 °C
Final temperature = 78 °C
The standard heats of formation
−285.8 kJ/mol H2O(l)
−924.54 kJ/mol Mg(OH)2(s)
Step 2: The equation
The heat is produced by the following reaction:
Mg(s)+2H2O(l)→Mg(OH)2(s)+H2(g)
Step 3: Calculate the mass of Mg needed
Using the standard heats of formation:
−285.8 kJ/mol H2O(l)
−924.54 kJ/mol Mg(OH)2(s)
Mg(s) + 2 H2O(l) → Mg(OH)2(s) + H2(g)
−924.54 kJ − (2 * −285.8 kJ) = −352.94 kJ/mol Mg
(4.184 J/g·°C) * (78 g) * (78 - 29)°C = 15991.248 J required
(15991.248 J) / (352940 J/mol Mg) * (24.3 g Mg/mol) = 1.1 g Mg
We need 1.1 grams of Mg
The formation of ammonia gas involves reacting hydrogen gas and nitrogen gas in a mole ratio of 3 to 1. as shown below:
<h3>What is the equation of the formation of ammonia?</h3>
Ammonia gas is formed from the reaction between nitrogen gas and hydrogen gas.
Three moles of hydrogen gas will react with 1 mole of nitrogen gas to form 2 moles of ammonia gas.
The equation of the reaction is given below as:

Therefore, the formation of ammonia gas involves reacting hydrogen gas and nitrogen gas in a mole ratio of 3 to 1.
Learn more about ammonia gas at: brainly.com/question/7982628
Answer:
Lengths. of. Naphthalene. Figure 3.20 shows that there are two equivalent ... all the carbon–carbon bonds of benzene are identical and are intermediate in length ... A typical carbon–carbon single bond has a length of 1.54 Å, and a double ... of how resonance can be used to explain or predict experimental observations.Explanation:
Answer:
After the solution is heated, but before additional solute is added
Explanation:
An unsaturated solution is a solution that contains less solute than it can normally hold at a given temperature. Hence an unsaturated solution can still dissolve more solute.
When the solution is heated, the saturated cold solution becomes an unsaturated hot solution which is capable of dissolving more solute at this point.
Once more solute is dissolved, the solution becomes saturated again just before it begins to cool since no more solute dissolves in the solution at some point before cooling and addition of seed crystals.