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Chemistry
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Chlorine, 15 g, is contained in a vessel with a volume of 0.80 dm3 at 330 K. Calculate the pressure exerted when the chlorine is treated as a perfect (ideal) gas giving your answer in terms of kPa

PV=nRT V=0.8 dm3=0.8 x 10-3 m3P= nRT/V R= 8.31 J K-1mol-1 T= 330 K n= mass/mr n = 15 g/ (2x35.5 g mol-1) n = 0.212 mol P = (0.212 mol x 8....

Answered by Harry G. Chemistry tutor
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What is meant by equilibrium

When a reaction is in equilibrium, it means the rate of the forward reaction is equal to the rate of the backwards reaction. The concentrations of the reactants and products remain constant, but this does...

Answered by Celine Iswarya P. Chemistry tutor
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Reaction between a metal and acid gives?

Metal + Acid --> Metal salt + H2 (g)E.g. Cu + HCl --> CuCl2 + H2

Answered by Celine Iswarya P. Chemistry tutor
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The student did another experiment using 20 cm3 of sodium hydroxide solution with a concentration of 0.18 mol/dm3 . Relative formula mass (Mr) of NaOH = 40 Calculate the mass of sodium hydroxide in 20 cm3 of this solution.

Answer= 0.144gWorking out:mol= conc x volume (in dm^3)mass = mol x RFFSo,mol = 0.18 x (20/1000) = 0.0036molMass= 0.036 x 40 = 0.144g

Answered by Celine Iswarya P. Chemistry tutor
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ii) The maximum permitted sulfate concentration in water is 250mg dm^-3, 200cm^3 of aqueous BaCl2 is added to 300cm^3 of water at the maximum permitted sulfate level, and a white precipitate formed. Calculate the minimum conc. (mol dm^3)of the BaCl2

250x10^-3= 0.25g dm-3sulfate mr x moles = mass0.25 x 3/10 = 0.075 g sulfatemr sulfare = 32 + (4 x 16) = 96moles sulfate = 0.075/96 = 7.8125 x10-4Per equation above: moles SO4

Answered by Josh M. Chemistry tutor
1646 Views

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