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
1.0566 gal.
H O W E V E R
the constant formula you can use for solving and simplifying equations such as these, you can simply look up the metric system online and figure one out. hoped this helped!
Answer:
0.554M of Calcium Bromide
Explanation:
Molarity by defintion is #of moles of something/litres of solution.
Therefore, here, we have 0.277 moles of calcium bromide and 500mL (divide 500ml by 1000 to go from mL to L because for every 1L there's 1000mL) or 0.5L.
Molarity= 0.277/0.5 = 0.554M of Calcium Bromide
Answer:c which is 70
Explanation:
i took the test and got it right
2.91 mol Al * ( 26.982 g Al / 1 mol Al) = 78.518 grams
