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
Answer:
Aluminum lodide - 680 degrees
fructose- 824 degrees
potassium bromide- 2,615 degrees
calcium bromide 3, 515 degrees
lowest boiling point is 680 and highest is 3,515.
hope this helps:)
Answer:
A Winter storm usually does not come with a warm fronthey warm front.
Explanation:
When the front passes over an area it means to change in the weather. Many. Many friends cause weather events such as rain thunderstorms gusty winds tornadoes and hurricanes.
