Recent content by Madelin Pierce

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    Chemistry Molar solubility lab questions

    I’m confused on how to calculate molar solubility because I don’t see what’s the difference between that and Ksp.
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    Chemistry Lab Titration curve calculations

    Homework Statement Homework EquationsThe Attempt at a Solution I don’t have a clue on how to do the calculations. I know the equivalency point is at the steepest part of the derivative graph. Please don’t say just figure it out, I truly need help. I haven’t learned titrations in lecture yet...
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    Chemistry Equilibrium Partial Pressures

    I use the equation Kp=Kc(RT)^change in n. I plug in Kp=(.0415)(.08206)(629.15)^1= 2.1425. Then, I solve for PV=nRT for PCl5: P(2.00L)=(.100mol)(.08206)(629.15K), equaling 2.5814atm. I use Kp= Products/Reactants which is 2.1425=x^2/(2.5814-x) where the Xs are from an ice table. I was told by my...
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    Chemistry Equilibrium Partial Pressures

    Homework Statement Kc = 4.15 x 10-2 at 356°C for PCl5(g) ↔ PCl3(g) + Cl2(g). A closed 2.00 L vessel initially contians 0.100 mol PCl5. Calculate the total pressure in the vessel (in atm to 2 decimal places) at 356°C when equilibrium is achieved. Homework Equations PV=nRT Kp= Kc(RT)^change in...
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    Kc = 1.88 for 2HI(g) ↔ H2(g) + I2(g)

    I know the reaction reversed and doubled. But I don’t know how to solve for Kc mathematically with this set-up.
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    Kc = 1.88 for 2HI(g) ↔ H2(g) + I2(g)

    Homework Statement Kc = 1.37 for HI(g) ↔ 1/2 H2(g) + 1/2 I2(g). What is Kc (to 3 decimal places) for H2(g) + I2(g) ↔ 2HI(g)? Homework Equations Kc= Products/Reactants, Ice Tables The Attempt at a Solution I tried an ice table, but I’m not sure what to do with the Kc value I’m given when I...
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    Rate constant calculation Using the Arrhenius equation

    Homework Statement A reaction has a rate constant of 0.0117/s at 400.0 K and 0.689/s at 450.0 K. What is the value of the rate constant (to 1 decimal place) at 538 K? Homework Equations ln k= ln A-E[a]/RT The Attempt at a Solution I'm not sure how to approach this problem, although I know...
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    Chemistry: dissolving aluminum metal in hydrochloric acid

    Homework Statement Aluminum metal dissolves in hydrochloric acid. What volume, in mL, of 1.58 M HCl is needed to completely dissolve 3.200 g of aluminum?(Hint: write the single-replacement reaction first) Homework Equations V[1]M[1]=V[2]M[2] The Attempt at a Solution I got the...
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    Balancing Aqueous Calcium Nitrate & Ammonium Phosphate

    Homework Statement Problem: Predict the products between aqueous calcium nitrate and aqueous ammonium phosphate. Homework EquationsThe Attempt at a Solution 3Ca(NO[3])[2](aq) + 2(NH[4])[3]PO[4](aq) = Ca[3](PO[4])[2] + 6(NH[4])(NO[3]) I'm not sure what the states of matter would be on the...
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    Torque and counterclockwise versus clockwise

    Homework Statement How do I determine whether something will go counterclockwise or clockwise? I know the hands of the clock idea, but how do I know in a torque problem? Also, how do I know what force will bring something into equilibrium? Homework EquationsThe Attempt at a Solution
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    Heat Energy and Power -- Heating ice in a copper pan

    Homework Statement Calculate the power needed to heat a 1360 gram chunk of ice in a 600 gram copper pan from 0 ◦C to the system’s final temperature of 35◦C in a time of 10 minutes. Homework Equations Q=mcchange in temp Q=mLf P= energy/time The Attempt at a Solution I thought that adding the...
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    Calculate Avg Power Output for 75kg Person Walking 100ft in 3min

    Kinetic energy would be zero because the problem indicates the word stop as in rest. Potential energy is there as the person is higher above ground than before. I do think it's a P= energy/time problem
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    Calculate Avg Power Output for 75kg Person Walking 100ft in 3min

    Homework Statement Calculate average power output for a 75 kilogram person walking from ground level up stairs to a final height of 100 feet (then stops) in a time of 3 minutes Homework Equations P= work/ time or P=Fv The Attempt at a Solution I thought this was more of an energy problem...
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