MedVision ad

Search results

  1. R

    Help With Conquering Chemistry Titration Questions

    Data: 5.267g Na2CO3 (with mm = 105.81) moles = mass/mm -> 5.267/105.81 = 0.0497... moles of Na2CO3 moles = conc*volume (in L) -> 0.0467... = C * 0.25 (250mL) -> 0.1991...mol/L or M 2HCl + Na2CO3 -> H2O + CO2 + 2NaCl no moles of HCl = 2 * no moles of Na2CO3 C * V (HCl) = 2 * (C * V) (Na2CO3) C *...
  2. R

    Biology practical assessment task! Mab

    If it's a practical worth 30% then I'd assume it's something large and that's most likely going to be 'assess the effect of substrate concentration, pH and temperature on the activity of enzymes'. There's the enzyme that is in the liver, and you have to cut up a bit of liver and put it in a test...
  3. R

    Water concentration affecting pH??

    The amount of CO2 is proportional to the increase of acidity in water. That is, the more CO2 in the water, more acidic the solution becomes and the lower the pH That is because CO2 dissolves in water to produce carbonic acid which increasing the pH In terms of this biologically, the blood is...
  4. R

    Needing Assistance With Gearing :)

    It's not always bad to have a high % of gearing debt to equity. It depends on the businesses position. If it is in start up or expansion it is likely to have a higher % than if the business has settled in or at a steady state. But in saying that, yes, it is generally bad to have a high % as it...
  5. R

    QANTAS case study -Help

    Well for a) it may be referring to the classification of the business when it's talking about type. That is, "QANTAS is a large public company operating globally in the tertiary sector" For b) is this year 11 work or HSC, because I have no idea if it is in the HSC, but for preliminary then it...
  6. R

    Extremely off result for titration

    Those answers are fine, just take an average and talk about it in your discussion. It would be weird if they were 15 or 20 mL out, 2 or 3 is fine.
  7. R

    Titration calculation problem

    n = C.V (in litres) Moles of HCl = 0.1 times 0.5 = 0.05 moles Moles of NaOH = 0.05 times 0.5 = 0.025 moles HCl will be in excess by 0.025 moles (HCl - NaOH) This means there will be 0.025 moles of H+ ions left over (as opposed to left of OH- ions if the NaOH was in excess.) The total volume...
Top