vanadium has an atomic mass of 50.9415 amu. it has two common isotopes.one isotopes has a mass of 50.9440 amu and a relative abu

Answer: The correct choice is (3).

Explanation:

The atomic number for lithium is 3, and its electron configuration is 2, 1. To achieve stability, it will lose one electron, thus forming a single bond.

Chlorine has an atomic number of 17, with 7 electrons in its outer shell. It will gain an electron to reach stability, resulting in the formation of a single bond.

Nitrogen, with an atomic number of 7 and an electron distribution of 2, 5, must acquire 3 electrons to stabilize. Therefore, nitrogen can form both triple and double bonds.

Hydrogen's atomic number is 1, and it stabilizes by gaining one electron, always forming a single bond.

Fluorine has an atomic number of 9 with 2, 7 electron distribution. It needs to gain one more electron to complete its octet, so it also forms a single bond.

Thus, we conclude that among the listed options, nitrogen is most capable of forming multiple (double or triple) bonds.

Answer: The required mass of KNO₃ is,

Explanation: Provided,

Molal-freezing-point-depression constant for water =

Amount of water = 275 mL

Molar mass of KNO₃ = 101.1 g/mole

First, we need to determine the mass of the water.

Density of water = 1.00 g/mL

Next, we will figure out the mass of KNO₃

Formula utilized:

where,

= change in freezing point

= freezing point of solution =

= freezing point of water =

i = Van't Hoff factor = 2  (for KNO₃ electrolyte)

= freezing point constant for water =

m = molality

Substituting all known values into this equation yields

Consequently, the mass of KNO₃ that must be added is,

The Δ H value for butane (g) is -124.7 kJ/mol.

The Δ H value for CO2 (g) is -393.5 kJ/mol.

The Δ H value for H2O (g) is -241.8 kJ/mol.

The mass of butane is 8.30 grams.

Butane has a molar mass of 58 g/mol.

Considering the reaction,

C₄H₁₀ + 6.5 O₂ = 4CO₂ + 5H₂O

To determine the Δ H° of the reaction:

ΔH°rxn = ∑nH° f (products) - ∑nH° f (reactants)

By substituting values, we find that

Δ H° rxn = 4 (-393.5) + 5 (-241.8) - (-124.7)

= -1574 -1209 + 124.7

= -2783 - 124.7

= -2658.3 kJ/mol

Now, we will calculate how many moles of butane are in 8.30 grams.

Number of moles = mass/molar mass

= 8.30 / 58

= 0.143 moles

Therefore, the total energy released during the reaction is given by,

Q = number of moles × ΔH° rxn

= 0.143 × (2658.3)

= 380.14 kJ

Thus, the total heat released in the reaction is 380.14 kJ.