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Chemistry help

for 9, fine the # of moles of each of the two compounds, then you'll have the moles of na and cl. but you want to find the limiting reagent and use that as the molar quantity because its a 1:1 ratio of na to cl. finally take the number of moles and multiply it to find the mass of the nacl.
 
No offence, but if you have no idea how to solve these problems, you need to ask your teacher, not some random internet forum. Number 20, especially, are as basic as it gets.

Edit, sorry, you apparently do these problems without a teacher.
 
for 19 do the negative log of 1.5. that'll give you the concentration of H+ ions. i believe hno3 dissociates completely, so whatever that concentration is, the total quantity of H+ and by the 1:1 ratio, is also the quantity of Hn03
 
20.) Products are Na2SO4 + H20. Ready, set, balance.

Edit: I'm slower than molasses. But you might be, too.
 
Originally posted by: Inspector Jihad
for 19 do the negative log of 1.5. that'll give you the concentration of H+ ions. i believe hno3 dissociates completely, so whatever that concentration is, the total quantity of H+ and by the 1:1 ratio, is also the quantity of Hn03
Click to expand...


Thanks everyone.

Ok, i took the -log but what happens to the 300mL? It doesn't play a role?
 
God I hate chemistry...
Don't you have to find the molar mass first here? Or am I wrong?
EDIT: Pssh, nvm I suck at Chem.
 
Originally posted by: The Godfather
Originally posted by: Inspector Jihad
for 19 do the negative log of 1.5. that'll give you the concentration of H+ ions. i believe hno3 dissociates completely, so whatever that concentration is, the total quantity of H+ and by the 1:1 ratio, is also the quantity of Hn03
Click to expand...


Thanks everyone.

Ok, i took the -log but what happens to the 300mL? It doesn't play a role?
Click to expand...

the number you'll calculate will be the amt of solution in 1 liter, you jsut want 300 ml
 
Originally posted by: Inspector Jihad
Originally posted by: The Godfather
Originally posted by: Inspector Jihad
for 19 do the negative log of 1.5. that'll give you the concentration of H+ ions. i believe hno3 dissociates completely, so whatever that concentration is, the total quantity of H+ and by the 1:1 ratio, is also the quantity of Hn03
Click to expand...


Thanks everyone.

Ok, i took the -log but what happens to the 300mL? It doesn't play a role?
Click to expand...

the number you'll calculate will be the amt of solution in 1 liter, you jsut want 300 ml
Click to expand...


i got -0.18 after doing the -log.. then i multiply by 0.300 L right?
 
#19:

pH= - log [ H ]

pH= 1.5
--> 1.5= - log [ H ]

-1.5= log [ H ]

[ H ]= 0.036 M, or a 36 mM HNO3 soln. Molec Weight of HNO3 is 32 g/mol

0.036 mol/L (M) * 0.30L= 0.0108 mol

0.0108 mol * 32g/mol = 0.3456g

Did some rounding, so check the math. This is quite easy. pH is an expression of H concentration. The (easy) trick of this is finding the [ H ]. The rest is unit conversion.
This makes 9 look super easy.

#9:
0.05L * 3M= 0.15 mol. This is true for both the acid and the base. Just figure out the molec weights of each, mult by the mols to get grams, then add the grams up. I'll let you do this.😉:evil:


#20: already answered








 
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