mass of pentane : = 30.303 g
moles of Al₂(CO₃)₃ : = 0.147
<h3>Further explanation</h3>
Given
1. Reaction
C₅H₁₂+8O₂→6H₂O+5CO₂.
45.3 g water
2. 2AlCl₃ + 3MgCO₃ → Al₂(CO₃)₃ + 3MgCl₂
37.2 MgCO₃
Required
mass of pentane
moles of Al₂(CO₃)₃
Solution
1. mol water = 45.3 : 18 g/mol = 2.52
From equation, mol ratio of C₅H₁₂ : H₂O = 1 : 6, so mol pentane :
= 1/6 x mol H₂O
= 1/6 x 2.52
= 0.42
Mass pentane :
= mol x MW
= 0.42 x 72.15 g/mol
= 30.303 g
2. mol MgCO₃ : 37.2 : 84,3139 g/mol = 0.44
mol Al₂(CO₃)₃ :
= 1/3 x mol MgCO₃
= 1/3 x 0.44
= 0.147
Answer: try to understand coz the question is not valid
Explanation: Explain the relationship between forward and reverse reactions at equilibrium and predict how changing the amount of a reactant or product (creating a stress) will affect that relationship.For example (select one from each underlined section)If the amount of (reactant or product) increases, the rate of the (forward or reverse)reaction will (increase or decrease)to reach a new equilibrium. If the amount of (reactant or product) decreases, the rate of the (forward or reverse)reaction will (increase or decrease)to reach a new equilibrium. Procedure: Access the virtual lab and complete the inquiry experiment
Answer:
1°C temperature change will be observed if a sample of 100 g of ethylene glycol antifreeze solution.
Explanation:
Mass of ethylene glycol = m = 100 g
Specific heat capacity of ethylene glycol = c = 3.5 J/g°C
Change in temperature of ethylene glycol = ΔT
Heat loss by the ethylene glycol = Q = 350 J
ΔT = 1°C
1°C temperature change will be observed if a sample of 100 g of ethylene glycol antifreeze solution.
Answer:
Percent error = 1.5%
Explanation:
Given data:
Measured value of density of graphite = 2.3 g/cm³
Percent error = ?
Solution:
Formula:
Percent error = [Measured value - Actual value / actual value] × 100
Actual/accepted value of density of graphite = 2.266 g/cm³
Now we will put the values:
Percent error = [2.3 g/cm³ - 2.266 g/cm³ / 2.266 g/cm³] × 100
Percent error = [0.034 g/cm³ / 2.266 g/cm³] × 100
Percent error = 0.015 × 100
Percent error = 1.5%
Answer:
111.15 g are required to prepare 500 ml of a 3 M solution
Explanation:
In a 3 M solution of Ca(OH)₂ there are 3 moles of Ca(OH)₂ per liter solution. In 500 ml of this solution, there will be (3 mol/2) 1.5 mol Ca(OH)₂.
Since 1 mol of Ca(OH)₂ has a mass of 74.1 g, 1.5 mol will have a mass of
(1.5 mol Ca(OH)₂ *(74.1 g / 1 mol)) 111.15 g. This mass of Ca(OH)₂ is required to prepare the 500 ml 3 M solution.
