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:
The answer to your question is 88.7 ml
Explanation:
Data
Volume = ?
Concentration of NaOH = 0.142 M
Volume of H₂C₄H₄O₆ = 21.4 ml
Concentration of H₂C₄H₄O₆ = 0.294 M
Balanced chemical reaction
2 NaOH + H₂C₄H₄O₆ ⇒ Na₂C₄H₄O₆ + 2H₂O
1.- Calculate the moles of H₂C₄H₄O₆
Molarity = moles/volume
Solve for moles
moles = Molarity x volume
Substitution
moles = 0.294 x 21.4/1000
Result
moles = 0.0063
2.- Use proportions to calculate the moles of NaOH
2 moles of NaOH ------------------ 1 moles of H₂C₄H₄O₆
x ------------------ 0.0063 moles
x = (0.0063 x 2) / 1
x = 0.0126 moles of NaOH
3.- Calculate the volume of NaOH
Molarity = moles / volume
Solve for volume
Volume = moles/Molarity
Substitution
Volume = 0.0126/0.142
Result
Volume = 0.088 L or 88.7 ml
Answer:
<h3>Sand cannot mixed on water not float on water .</h3>
<h3>When we mix sand and water , No reaction take place . The sand simply settles down at the bottom of the water container . This is why sand is heavier than water and therefore cannot float in water .</h3>
<h2>Hope this helps you ✌️</h2>
