Answer:
C. AlCl3 is a covalent compound.
Explanation:
It is formed by mutual sharing of electrons between two atoms each contributing equal number of electrons to the electron pair.
ΔHrxn = ΣδΗ(bond breaking) - ΣδΗ(bond making)
Bond enthalpies,
N ≡ N ⇒ 945 kJ mol⁻¹
N - Cl ⇒ 192 kJ mol⁻¹
Cl - Cl⇒ 242 kJ mol⁻¹
According to the balanced equation,
ΣδΗ(bond breaking) = N ≡ N x 1 + Cl - Cl x 3
= 945 + 3(242)
= 1671 kJ mol⁻¹
ΣδΗ(bond making) = N - Cl x 3 x 2
= 192 x 6
= 1152 kJ mol⁻¹
δHrxn = ΣδΗ(bond breaking) - ΣδΗ(bond making)
= 1671 kJ mol⁻¹ - 1152 kJ mol⁻¹
= 519 kJ mol⁻¹
Answer:
Explanation: Mendeleev arranged the elements on the basis of their atomic mass. Melting and boiling point were used as the physical characteristics in deciding the position of elements. He arranged the elements and wrote the formula of their oxides and hydrides which seemed to possess same chemical formula.
Explanation:
Simply put, density is how tightly “stuff” is packed into a defined space.
For example, a suitcase jam-packed with clothes and souvenirs has a high density, while the same suitcase containing two pairs of underwear has low density. Size-wise, both suitcases look the same, but their density depends on the relationship between their mass and volume.
Mass is the amount of matter in an object.
Volume is the amount of space that an object takes up in three dimensions.
Density is calculated using the following equation: Density = mass/volume or D = m/v.
If something is heavy for its size, it has a high density. If an object is light for its size it has a low density.
The relative densities of an object and the liquid it is placed in determine whether that object will sink or float.
The volume of the 0.15 M LiOH solution required to react with 50 mL of 0.4 M HCOOH to the equivalence point is 133.3 mL
<h3>Balanced equation </h3>
HCOOH + LiOH —> HCOOLi + H₂O
From the balanced equation above,
The mole ratio of the acid, HCOOH (nA) = 1
The mole ratio of the base, LiOH (nB) = 1
<h3>How to determine the volume of LiOH </h3>
- Molarity of acid, HCOOH (Ma) = 0.4 M
- Volume of acid, HCOOH (Va) = 50 mL
- Molarity of base, LiOH (Mb) = 0.15 M
- Volume of base, LiOH (Vb) =?
MaVa / MbVb = nA / nB
(0.4 × 50) / (0.15 × Vb) = 1
20 / (0.15 × Vb) = 1
Cross multiply
0.15 × Vb = 20
Divide both side by 0.15
Vb = 20 / 0.15
Vb = 133.3 mL
Thus, the volume of the LiOH solution needed is 133.3 mL
Learn more about titration:
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