Answer:
![CuSO_4 (aq) + Fe (s)\rightarrow FeSO_4 (aq) + Cu (s)](https://tex.z-dn.net/?f=CuSO_4%20%28aq%29%20%2B%20Fe%20%28s%29%5Crightarrow%20FeSO_4%20%28aq%29%20%2B%20Cu%20%28s%29)
Explanation:
Identify all of the compounds present:
- aqueous copper(II) sulfate: we have copper charged as 2+, sulfate anion has a charge of 2-. This means the charges are balanced if we take 1 ion of each to give
; - solid iron metal: iron metal is Fe, then including the state of iron, we have
; - aqueous iron(II) sulfate: iron has a charge of 2+, sulfate has a charge of 2-, so the positive charge balances the negative charge to give a formula of
; - solid copper metal: copper metal is Cu, then including the state of copper, we have
.
We then have a reaction:
![CuSO_4 (aq) + Fe (s)\rightarrow FeSO_4 (aq) + Cu (s)](https://tex.z-dn.net/?f=CuSO_4%20%28aq%29%20%2B%20Fe%20%28s%29%5Crightarrow%20FeSO_4%20%28aq%29%20%2B%20Cu%20%28s%29)
First you need to balance the equation. Ba(OH)2+2HNO3=Ba(NO3)2+H2O. Then you can get the ratio of mole number of the reactants. The ratio is Ba(OH)2:HNO3=1:2. So the molarity of Ba(OH)2 is 38.5*0.85/2/20=0.82 molar.
Answer:
False, on the celsius scale it is -273.15°
Explanation:
Answer:
37.5 mL
Explanation:
To solve this problem we can use the formula:
C₁ = 0.400 M
V₁ = ?
C₂ = 0.150 M
V₂ = 100 mL
So now we <em>solve for V₁</em>:
- 0.400 M * V₁ = 0.150 M * 100 mL
So we need 37.5 mL of 0.400 M CuSO₄(aq) to make 100 mL of 0.150 M CuSO₄(aq). That volume could be measured in the 50 mL buret.
Well. NaOH is a base. That's the first thing you need to watch for.
So to find the pOH, you take -log(.0001)
that would be 4. So now you have the pOH and <u>you still need to find the pH
</u>To find pH from pOH, you take 14(the maximum pH,sorta)-pOH(in this case 4)
14-4=10 The pH of NaOH is 10