The balanced chemical reaction is:
<span>Ca + Cl2 = CaCl2
</span>
We are given the amount of calcium metal to be used for this reaction. This will be the starting point for the calculations.
56 g Ca ( 1 mol Ca / 40.08 g Ca) (1 mol Cl2 / 1 mol Ca) ( 22.414 L Cl2 / 1 mol Cl2 ) = 31.32 L Cl2 gas produced from the reaction
Answer:
D. 450 J/kgK
Explanation:
Using the formula as follows:
Q = m × c × ∆T
Where:
Q = amount of heat absorbed/released (J)
m = mass of substance (g)
c = specific heat capacity
∆T = change in temperature (°C)
According to the information provided in this question:
Q = 267.3 kJ = 267300J
m = 18kg
∆T = 318K - 285K = 33K
c = ?
Q = m × c × ∆T
c = Q ÷ m∆T
c = 267300 ÷ 18 × 33
c = 267300 ÷ 594
c = 450 J/kgK
Answer:
1,033.56 grams of carbon dioxide was emitted into the atmosphere.
Explanation:
Energy absorbed by pork,E =
(assuming)
Total energy produced by barbecue = Q
Percentage of energy absorbed by pork = 10%


Since, it is a energy produced in order to indicate the direction of heat produced we will use negative sign.
Q = 
Moles of propane burnt to produce Q energy =n


According to reaction , 1 mol of propane gives 3 moles of carbon dioxide. then 7.83 moles of will give:
carbon dioxide gas.
Mass of 23.49 moles of carbon dioxide gas:
23.49 mol × 44 g/mol =1,033.56 g
1,033.56 grams of carbon dioxide was emitted into the atmosphere.
Answer :
The correct answer is %IC = 10 % and bond is covalent bond with slight polarity.
<u>Percent Ionic Character :</u>
It is defined as percent of ionic character present in a polar covalent bond . The formula of % ionic character (%IC) is given as follows :

Where Xa = Electronegativity of A atom and Xb = Electronegativity of B atom
Given : Molecule is TiAl₃
Electronegativity of Ti = 2.0
Electronegativity of Al = 1.6 ( From image shared )
Plug the value in above formula :



Value of e⁻¹ = 0.90
Percent ionic character = 1 - 0.90 * 100
Percent Ionic character = 10 %
<u>Since the % IC is 10 % , which is very less comparatively , hence the bond is covalent and very less polar .</u>