When 500.0 g of water is decomposed by electrolysis and the yield of hydrogen is only 75.3%, how much hydrogen chloride can be m

The formula for Molarity is given by:

                                  M = moles / V
To isolate V,
                              V = moles / M ------------------(1)
Moles can also be calculated as:
                                  moles = mass / M.mass -------------(2)
Substituting the value of moles from equation 2 into equation 1 yields:
                                  V = (mass / M.mass) / M
Plugging in the numbers gives:
                                  V = (45 g / 164 g/mol) / 1.3 mol/dm³

                                       V = 0.21 dm³.

Answer:

The calculated density of the liquid is 1470.43 kg/m³.

Explanation:

Given:

Mass of the solid sphere (m) = 6.1 kg

Density of the metal = 2600 kg/m³

To find the volume of the liquid:

Volume of the sphere can be calculated as 6.1 kg / 2600 kg/m³ = 0.002346 m³.

According to Archimedes' principle, the volume of water displaced is equal to the volume of the sphere.

Volume displaced = 0.002346 m³

The buoyant force formula is:

Where:

is the fluid's density,

g represents the acceleration due to gravity,

V indicates the volume displaced.

Referencing the free-body diagram of the sphere shown in the image:

Acceleration due to gravity = 9.81 ms⁻²

Tension force = 26 N

Using these in the equation to ascertain the liquid density yields:

Thus, the density of the liquid = 1470.43 kg/m³

Answer:

The percent yield of Br₂ in this reaction amounts to 96.15%

Explanation:

The reaction's balanced stoichiometric equation is:

2 NaBr + 1 Cl₂ → 2 NaCl + 1 Br₂

To calculate the percent yield:

Percent yield = 100% × (Actual yield)/(Theoretical yield)

To determine the theoretical yield:

5.29 g of NaBr reacts with an excess of chlorine; therefore, NaBr is the limiting reagent, controlling the possible yield of products.

We convert 5.29 g of NaBr to moles.

Number of moles = (Mass)/(Molar mass)

Molar Mass of NaBr = 102.894 g/mol

Number of moles = (5.29/102.894) = 0.0514121329 = 0.05141 mole

According to the stoichiometry of the reaction:

2 moles of NaBr yield 1 mole of Br₂

Thus, 0.05141 mole of NaBr will produce (0.05141×1/2) mole of Br₂, which is 0.0257 mole of Br₂

Theoretical yield = Expected mass of Br₂ from the reaction

= (Number of moles) × (Molar mass)

Molar mass of Br₂ = 159.808 g/mol

Theoretical yield of Br₂ = 0.0257 × 159.808 = 4.108 g

Calculating the percent yield:

Percent yield = 100% × (Actual yield)/(Theoretical yield)

Actual yield = 3.95 g

Theoretical yield = 4.108 g

Percent yield = 100% × (3.95/4.108) = 96.15%

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