Answer:
C)There would have been four pots in both cases
Explanation:
In an incomplete reaction of naphthalene and 2-bromo-2-methylpropane, we started with 2 spots (one each for the two starting materials) and we should end up with 3 spots If spots will appear clearly, one for naphthalene, one for the alkyl halide and one for the product. However, this is possible to have same number of spots initially and finally when we consider 2 spots each for naphthalene and 2-bromo-2-methylpropane then after reaction we have 2 spots of products as each 1 spot of reactants decreases. Finally making initial 4 and final 2+1+1 =4.
So, the answer is (C)
The amount of Na₂SO₄ would be 15.09 grams
<h3>Stoichiometric calculation</h3>
From the equation of the reaction:
4NaOH + 2H₂SO₄ --> 2Na₂SO₄ + 4H₂O
Mole ratio of NaOH and Na₂SO₄ = 2:1
Mole of 8.50 grams of NaOH = 8.50/40
= 0.21 mole
Equivalent mole of Na₂SO₄ = 0.21/2
= 0.11 mole
Mass of 0.11 mole Na₂SO₄ = 0.11 x 142.04
= 15.09 grams
More on stoichiometric reactions can be found here: brainly.com/question/8062886
Answer:
Products are stearate anion and water.
Explanation:
Stearic acid is a 18-carbon chain molecule containing -COOH group. IUPAC name of stearic acid is octadecanoic acid.
Molecular formula of stearic acid is 
.
When 
is added into stearic acid, removes a proton (
) from acidic -COOH group and forms stearate anion and water as products.
The balanced acid-base reaction is given as:
Structure of products are given below.
Answer:
Physical Change
Explanation:
hope this help brainliest pls
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
