Answer :
The percentage ionic character (%IC) equals 10%, indicating the bond is mostly covalent with slight polarity.
Percent Ionic Character:
This reflects the fraction of ionic nature within a polar covalent bond. The formula for %IC (% ionic character) is:

Here, Xa is the electronegativity of atom A and Xb is that of atom B.
Given: The compound is TiAl₃.
Electronegativity of Ti = 2.0
Electronegativity of Al = 1.6 (as shown in the provided image)
Substitute these values into the formula:



The value of e⁻¹ equals 0.90.
Therefore, percent ionic character = (1 - 0.90) × 100
Percent Ionic Character = 10%
Because the % IC is only 10%, which is relatively low, the bond is classified as covalent with minimal polarity.
Hello!
density = 2.67 g/cm³
volume = 30.5 mL
Thus:
Mass = density * volume
Mass = 2.67 * 30.5
Mass = 81.435 g
Response:
9.9 ml of 0.200M NH₄OH(aq)
Reasoning:
3NH₄OH(Iaq) + FeCl₃(aq) => NH₄Cl(aq) + Fe(OH)₃(s)
What volume in ml of 0.200M NH₄OH(aq) will fully react with 12ml of 0.550M FeCl₃(aq)?
1 x Molarity of NH₄OH x Volume of NH₄OH Solution(L) = 2 x Molarity of FeCl₃ x Volume of FeCl₃ Solution
1(0.200M)(Volume of NH₄OH Soln) = 3(0.550M)(0.012L)
=> Volume of NH₄OH Soln = 3(0.550M)(0.012L)/1(0.200M) = 0.0099 Liters = 9.9 milliliters
Answer: 0.0007 moles of
are released when the temperature rises.
Explanation:
To determine the moles, we utilize the ideal gas law, as follows:

where,
P = gas pressure = 1.01 bar
V = gas volume = 1L
R = gas constant = 
- Calculated moles at T = 20° C
The gas temperature = 20° C = (273 + 20)K = 293K
Substituting values into the equation gives:

- Calculated moles at T = 25° C
The gas temperature = 25° C = (273 + 25)K = 298K
Substituting values into the equation gives:

- Released moles =

Therefore, 0.0007 moles of
are released when the temperature increases from 20° C to 25° C.
A mixture is created by dissolving 1.43 mol of potassium chloride (KCl) in 889 g of water. The concentration of KCl works out to be 1.61 molal.
The amount of KCl is 1.43 mol
Water weighs 889 g
The molality can be calculated using the formula:
molality = moles of solute divided by kilograms of solvent
Since 1 kg equals 1000 g, 889 g is 0.889 kg.
Therefore, m = 1.43/0.889 = 1.61 molal.