The 0.25 volume in liters of 1.0 M 
solution is required to provide 0.5 moles of 
(aq).
Calculation,
The Potassium carbonate dissolves as follows:
→ 
The mole ratio is 1: 2
It means, the 1 mole required to form 2 mole of (aq).
To provide 0.5 mole of (aq) = 1 mole ×0.5 mole /2 mole required by .
To provide 0.5 mole of (aq) ,0.25 mole required by .
The morality of = 1 M = number of moles / volume in lit
The morality of = 1 M = 0.25 mole/ volume in lit
Volume in lit = 0.25 mole / 1 M = 0.25 mole/mole/lit = 0.25 lit
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.86km is the correct answer
<u>Answer:</u> The volume of the gas is 178.76 L
<u>Explanation:</u>
The ideal gas equation is given as:
.......(1)
where
P = pressure of the gas = 675 torr
V = volume of gas = ?
n = number of moles of gas = 6.45 moles
R = Gas constant = 62.36 L.torr/mol.K
T = temperature of the gas = ![27^oC=[27+273]K=300K](https://tex.z-dn.net/?f=27%5EoC%3D%5B27%2B273%5DK%3D300K)
Putting values in equation 1, we get:
Hence, the volume of the gas is 178.76 L
longitudinal wave.... is the right answer
<h2>
<em># Good morning</em></h2>
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
