How would I do this? A beaker contains 50.0 mL of 0.25 M aluminum nitrate solution. What is the minimum volume of 0.2 M sodium s
1 answer:
The minimum volume of 0.2M sodium sulfide that will precipitate out aluminum from 50.0 mL, 0.25 M aluminum nitrate would be 0.094 L or 94 mL
<h3>Stoichiometric calculation</h3>From the equation of the reaction:
Mole ratio of Na2S and Al(NO3)3 = 3:2
Mole of 50.0 mL, 0.25 M Al(NO3)3 = 50/1000 x 0.25
= 0.0125 mole
Equivalent mole of Na2S = 3/2 x 0.0125
= 0.0188 mole
Volume of 0.2M, 0.0188 mole Na2S = 0.0188/0.2
= 0.094 L or 94 mL
More on stoichiometric calculations can be found here: brainly.com/question/8062886
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The volume (in liters) that the gas will occupy if the pressure is increased to 13.5 atm and the temperature is decreased to 15 °C is 15 L
From the question given above, the following data were obtained:
Initial pressure (P₁) = 8.5 atm
Initial volume (V₁) = 24 L
Initial temperature (T₁) = 25 °C = 25 + 273 = 298 K
Final pressure (P₂) = 13.5 atm
Final temperature (T₂) = 15 °C = 15 + 273 = 288 K
<h3>Final volume (V₂) =? </h3>- The final volume of the gas can be obtained by using the combined gas equation as illustrated below:
Cross multiply
298 × 13.5 × V₂ = 204 × 288
4023 × V₂ = 58752
Divide both side by 4023
Therefore, the final volume of the gas is 15 L
Learn more: brainly.com/question/25547148