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
Mass % of nitrogen = mass of nitrogen*100 / total mass
= 14*100 / (1+ 14 + 32)
= 14*100 / 47
= 29.7 %
Answer: False
Explanation:
4C2H6 + 7O2 --> 6H2O + 4CO2
8 Carbons on the reactant side, but 4 Carbons on the Product
24 Hydrogens on Reactant, 12 H on product
But Oxygen is balanced, 14 on each side
Answer:
Lithium selenide Lithium selenide (Li2Se) 12136-60-6 Dilithium selenide lithium selenidolithium
Molecular Weight 92.9 g/mol
Dates Modify 2020-11-15 Create 2005-08-08