When a 3.22 g sample of an unknown hydrate of sodium sulfate, na2so4 ⋅ h2o(s), is heated, h2o (molar mass 18 g) is driven off. T

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,

Answer:

Cuando dos átomos se acercan entre sí, se genera un compuesto al compartir pares de electrones que cada uno de los átomos aporta, permitiéndoles alcanzar los 8 electrones de valencia (octeto) en su capa externa.

Explanation:

La configuración electrónica del elemento puede escribirse de la siguiente manera;

1s²2s²2p⁴

La configuración electrónica dada es equivalente a la del oxígeno, por lo tanto, tenemos;

El número de electrones en la capa de valencia = 2 + 4 = 6 electrones

Por consiguiente, cada átomo necesita 2 electrones para completar sus 8 electrones (octeto) en la capa externa.

Al acercarse los dos átomos, reaccionan y se combinan para formar un compuesto al compartir 4 electrones, 2 de cada átomo, de modo que cada átomo obtenga 2 electrones adicionales en su órbita externa en el nuevo compuesto y así se logre la configuración estable de octeto para cada uno de los átomos en el compuesto recién formado.

Answer:

Below are the downsides of each electron configuration model:

1). Dot Structures - They consume more space and fail to convey the electron distribution within orbitals.

2). Arrow and line diagrams complicate electron counting and also take up excessive space.

3). Written Configurations do not illustrate electron distribution in orbitals, leading to possible errors in counting electrons.

Answer: refer to the explanation

Explanation:

This question requires calculating mass. To determine the mass, we need values for two parameters, namely the specifications for the grade 70 tow chain and the mass per unit length given.

Assuming the mass per unit length is 3 kilograms per meter (kg/m) and the length of the grade 70 tow chain is 5 meters (m).

Thus, the formula used to compute the mass of the chain is as follows;

Mass of the chain = mass per unit length (kg/m) × length ---------------------------------------------------------------------------------------------------------------------(1).

Mass of the chain = 3 kg/m × 5 m.

Mass of the chain equals 15 kg.

Answer:

CaS, CaBr₂, VBr₅, and V₂S₅.

Explanation:

• The ionic compound must exhibit neutrality; its total charge should equal zero.

• A binary ionic compound is formed from two distinct ions.

Ca²⁺ combines with either Br⁻ or S²⁻ to create binary ionic compounds.

• CaS is created when Ca²⁺ pairs with S²⁻ resulting in the neutral binary ionic compound CaS.

• CaBr₂ results from the combination of one mole of Ca²⁺ with two moles of Br⁻ to form the neutral binary ionic compound CaBr₂.

V⁵⁺ can also unite with either Br⁻ or S²⁻ to produce binary ionic compounds.

• V₂S₅ is formed when two moles of V⁵⁺ bond with five moles of S²⁻ yielding the neutral binary ionic compound V₂S₅.

• VBr₅ is produced by combining one mole of V⁵⁺ with five moles of Br⁻ to form the neutral binary ionic compound VBr₅.

Thus, the empirical formulas for four binary ionic compounds that may be produced are: CaS, CaBr₂, VBr₅, and V₂S₅.