Answer:
1.03 M
Explanation:
Step 1: Write the balanced equation
NaOH + HCl ⇒ NaCl + H₂O
Step 2: Calculate the reacting moles of HCl
30.0 mL (0.0300 L) of 0.500 M HCl react.
0.0300 L × 0.500 mol/L = 0.0150 mol
Step 3: Calculate the moles of NaOH that react with 0.0150 moles of HCl
The molar ratio of NaOH to HCl is 1:1. The moles of NaOH that react are 1/1 × 0.0150 mol = 0.0150 mol.
Step 4: Calculate the molar concentration of NaOH
0.0150 moles of NaOH are in 14.5 mL (0.0145 L).
M = 0.0150 mol/0.0145 L = 1.03 M
<h2>Answer:</h2>
Arrangement of inter molecular forces from strongest to weakest.
- Hydrogen bonding
- Dipole-dipole interactions
- London dispersion forces.
<h3>Explanation:</h3>
Intermolecular forces are defined as the attractive forces between two molecules due to some polar sides of molecules. They can be between nonpolar molecules.
Hydrogen bonding is a type of dipole dipole interaction between the positive charge hydrogen ion and the slightly negative pole of a molecule. For example H---O bonding between water molecules.
Dipole dipole interactions are also attractive interactions between the slightly positive head of one molecule and the negative pole of other molecules.
But they are weaker than hydrogen bonding.
London dispersion forces are temporary interactions caused due to electronic dispersion in atoms of two molecules placed together. They are usually in nonpolar molecules like F2, I2. they are weakest interactions.
The equilibrium constant, Kc=0.026
<h3>Further explanation</h3>
Given
1.72 moles of NOCI
1.16 moles of NOCI remained
2.50 L reaction chamber
Reaction
2NOCI(g) = 2NO(g) + Cl2(g).
Required
the equilibrium constant, Kc
Solution
ICE method
2NOCI(g) = 2NO(g) + Cl2(g).
I 1.72
C 0.56 0.56 0.28
E 1.16 0.56 0.28
Molarity at equilibrium :
NOCl :
NO :
Cl2 :
![\tt Kc=\dfrac{[NO]^2[Cl_2]}{[NOCl]^2}\\\\Kc=\dfrac{0.224^2\times 0.112}{0.464^2}=0.026](https://tex.z-dn.net/?f=%5Ctt%20Kc%3D%5Cdfrac%7B%5BNO%5D%5E2%5BCl_2%5D%7D%7B%5BNOCl%5D%5E2%7D%5C%5C%5C%5CKc%3D%5Cdfrac%7B0.224%5E2%5Ctimes%200.112%7D%7B0.464%5E2%7D%3D0.026)
