Answer:
beakers and flasks
Explanation:
its better to use accurate measuring instruments for measuring volume
Answer:
Chlorine is limiting reactant
Explanation:
Based on the reaction:
Cl₂ + 2NaOH → NaClO + NaCl + H₂O
<em>1 mole of chlorine reacts with 2 moles of NaOH</em>
<em />
To find limiting reactant, we need to determine the moles of the reactants:
<em />
<em>Moles Cl₂ -Molar mass: 70.9g/mol-:</em>
800lb Cl₂ * (453.6g / 1lb) * (1mol / 70.90g) =
5118 moles Cl₂
<em>Moles NaOH -Molar mass: 40g/mol-:</em>
1200lb NaOH * (453.6g / 1lb) * (1mol / 40g) =
13608 moles NaOH
For a complete reaction of 13608 moles of NaOH you need:
13608 moles NaOH * (1mol Cl₂ / 2 moles NaOH) = 6804 moles of Cl₂
As the solution contains just 5118 moles of chlorine,
<h3>Chlorine is limiting reactant</h3>
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
Explanation:If the mass of the object stays the same but the volume of the object decreases then its density becomes greater. If the volume of the object stays the same but the mass of the object increases then its density becomes greater.
Answer:
Na(OH)4
Explanation:
Look at the charges and add them up
Na(OH) Na(OH)