Question

In: Chemistry

A)assuming volumes are additive, what volume of water must be added to 28.0 mL of 1.25...

A)assuming volumes are additive, what volume of water must be added to 28.0 mL of 1.25 M FeCl3 to prepare a 0.650 M FeCl3 solution?
B) what mass of Ca(OH)2 would be required to neutralize 120.0 mL of 0.350 M HClO4? molar mass= Ca(OH)2 74.08; HClO4 100.46
C) what is the minimum volume of 0.08475 M KOH needed to neutralize 10.00 mL of 0.1801 M H2C2O4?
D) a 10.0 L sample of an ideal gas at 25.0°C is heated at constant pressure until the volume has doubled. what is the final temp?

Solutions

Expert Solution

A) dilution formula is M1V1= M2V2

M1 = 1.25 M, M2 = 0.65 M,

V1 = 28 mL, V2 = ?

V2 = M1V1/M2 = (28 . 1.25)/0.65

V2 = 53.85 mL needed to add.

B) 2 HClo₄ + Ca(OH)₂ = Ca(Clo₄)₂ + 2 H₂O

here, one mole of ca(OH)2 requied to quench two moles of HClO4

no.of moles of HClO4 = molarity mltiply with volume in lit.

Given that, moles of HClO4 = 0.35 . 0.12 = 0.042 moles

so, 0.21 moles of Ca(OH)2 is requied to neutralize 0.42 moles of HClO4.

mass of Ca(OH)2 is equals to 0.21 multiply by74.08.

i.e., 15.56 g of Ca(OH)2 needed.

C) H2C2O4.2H2O + 2KOH = K2C2O4 + 4H2O

no.of moles of H2C2O4 = 0.1801 multiply by 0.01 lt. = 0.001801 moles

two moles of KOH is required per each mole of K2C2O4.

i.e., volume = 0.08475/0.003602 = 25.52 lt.

D) according to charles law, If the pressure of a gas sample is kept constant, the volume of the sample will vary directly with the temperature in Kelvin (Figure 9.9). As the temperature increases, so will the volume; if the temperature decreases, the volume will decrease. This relationship can be expressed by an equation relating the initial volume (V₁) and initial temperature (T₁ measured in K) to the final volume (V₂) and final temperature (T₂ measured in K). At constant pressure,

V₁V₂

T₁T₂

V1 = 10 L, V2 = 20 L, T1 = 298 K, T2 = ?

T2 = 2(298) = 596 K

queen_honey_blossom answered 1 year ago

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