genral chemistry help

[RING12]

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what is the molarityof 85% h3po4. if the density of the solution is 1.7 g/mol

A=1000*(1/1.70)*(100/15)*1.70/1


what is the equblirium constan K, at 400k if 3.00 moles of co2 introduced in to a 2 liter container 20% dissociated at eq? 2c02___________ 2co+o2

 

[desert]

what is the molarityof 85% h3po4. if the density of the solution is 1.7 g/mol

A=1000*(1/1.70)*(100/15)*1.70/1


what is the equblirium constan K, at 400k if 3.00 moles of co2 introduced in to a 2 liter container 20% dissociated at eq? 2c02___________ 2co+o2

1)

H3PO4= 3+31+(16*4) = 98 gr/mol
Density = mass/ volume =1.7 gr/L (I assume that you wrote the wrong unit here)

CM (H3PO4) (mol /L) = 85/100 * 1.7/1 (gr/L) * 1/98 (mol/gr)

2)
CO2: 3mol / 2Lit = 3/2 (mol/L)
CO :3/2 * 20/100 = 0.3 (mol/L)


2CO2 ----------- 2CO + O2
3/2 (mol/L) 0.3 (mol/L) 0.3 (mol/L)


k=[ CO]^2 [O2] /[ CO2]^2 = (0.3)^2 * (0.3) / (3/2)^2 = 0.012 = 1.2 * 10^2



Where did you got A=1000*(1/1.70)*(100/15)*1.70/1?

Desert

 

[RING12]

1)

H3PO4= 3+31+(16*4) = 98 gr/mol
Density = mass/ volume =1.7 gr/L (I assume that you wrote the wrong unit here)

CM (H3PO4) (mol /L) = 85/100 * 1.7/1 (gr/L) * 1/98 (mol/gr)

2)
CO2: 3mol / 2Lit = 3/2 (mol/L)
CO :3/2 * 20/100 = 0.3 (mol/L)


2CO2 ----------- 2CO + O2
3/2 (mol/L) 0.3 (mol/L) 0.3 (mol/L)


k=[ CO]^2 [O2] /[ CO2]^2 = (0.3)^2 * (0.3) / (3/2)^2 = 0.012 = 1.2 * 10^2



Where did you got A=1000*(1/1.70)*(100/15)*1.70/1?

Desert

i got them from dat archiver. may i ask you how did you get the 98 and 100. and what is cm means?

 

[desert]

i got them from dat archiver. may i ask you how did you get the 98 and 100. and what is cm means?

CM = molarity = mol / liter = mol/L
85% H3PO4 = 85 /100 H3PO4
Molecular weigth of H3PO4 = 3(1)+31+(16*4) = 98 gr/mol
Since H=1 gr/mol, P=31 gr/mol, O=16 gr/mol

Desert