physical science

posted by .

What is the Ratio for Al2(CrO4)3

  • physical science -

    2 Al ions to 3 CrO4 ions.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Intro to Physical Science

    Define physical science Andrew: I went to and typed in physical science. The following is what it returned.I hope this helps. physical science  1. any of the natural sciences dealing with inanimate matter or …
  2. Chemistry

    What molar [] of silver ion could exist in a solution in which the [] of CrO4^2- is 1.0*10^-4 (Ksp of Ag2CrO4= 1.1*10^-12) Ksp= [Ag]^2 [CrO4^2-]
  3. chemistry--URGENT, PLEASE HELP

    What concentration of silver chromate (Ksp = 9.0 x 10-12) will dissolve to make a saturated solution in water?
  4. physical science

    Calculate (to the nearest 0.1 u) the formula mass of these compounds. (a) rubidium nitrate, RbNO3 u (b) aluminum peroxide, Al2(O2)3 u (c) water, H2O u
  5. science (chem)

    consider the fo9rmula for Al2(SO4)3,which is used in perspirants: a. how many moles of sulfur are present i 3.0 moles of Al2(SO4)3 b. ow many moles of aluminum ions are present in 0.40 mole of Al2(SO4)3 for a and b i multiplied by …
  6. chemistry

    Hi, the question is: in each reaction, identify what have been oxidised and reduced. The equation is: 2Al + 3H2SO4 ---> Al2(SO4)3 + 3H2 I am confused about the Al2(SO4)3 I know what oxides and reductions us I just don't understant …
  7. chemistry

    Use the balanced equation below to answer the following questions. 2Al(NO3)3 + 3Na2CO3 Al2(CO3)3(s) + 6NaNO3 - What is the ratio of moles of Al(NO3)3 to moles Na2CO3?
  8. physical Science

  9. chemistry

    3.50 mol Al(NO3)3 reacts with 5.00 mols K2CrO4, the total number of moles of Al2(CrO4)3 and KNO3 would be?
  10. Chemistry

    Balance the reaction using half reactions under BASIC conditions: Al + CrO4^-2 ---> Al(OH)3 + Cr(OH)3 I tried to solve this question and ended up with this: Here are the two half equations Al ---> Al(OH)3 CrO4^-2 ---> Cr(OH)3 …

More Similar Questions