# Need help with some chemistry homework questions (1 Viewer)

#### Attachments

• 1.2 MB Views: 7

#### jimmysmith560

##### Le Phénix Trilingue
Moderator
Would the following help with the question in the first attachment?

$\bg_white Pb\left(s\right)+2AgNO_3\left(aq\right)\:\rightarrow \: Pb\left(NO_3\right)_2\left(aq\right)+2Ag\left(s\right)$

$\bg_white n_{Pb}=\frac{20.72\:g}{207.2\:g\:mol^{-1}}=0.1000\:mol$

$\bg_white n_{Ag}=0.100\:L\:\times 1.00\:mol\:L^{-1}=0.100\:mol$

$\bg_white \text{Mole ratio}\: Pb\::\:AgNO_3\:\text{is}\:1\::\:2$

$\bg_white \therefore 0.100\:mol\:AgNO_3\:\text{and}\:0.0500\:mol\: Pb\:\text{used}$

Moles of remaining ions and solids:

$\bg_white Pb^{2+}\left(aq\right)=0.0500\:mol$

$\bg_white Ag^+\left(aq\right)=0\:mol$

$\bg_white NO_3^{\:-}\left(aq\right)=0.100\:mol$

$\bg_white Pb\left(s\right)=0.0500\:mol$

$\bg_white Ag\left(s\right)=0.100\:mol$

#### Eagle Mum

##### Well-Known Member
@Dane Red, regarding comments in another thread, would any further explanation for Q1 be useful?

#### Eagle Mum

##### Well-Known Member
Q2 requires a little bit of general knowledge about the pH of various solutions (I don’t know if these are described as part of the curriculum).

HCl is a strong acid (red)
Ammonia is a relatively strong base when dissolved in water (yellow)
Lemon is a weak acid (orange)
Baking soda which is a weak base is commonly used as an oven cleaner (yellow)
Pure water is neutral (yellow)

if you arrange these in order of increasing pH:
Strong acid - red
Weak acid - orange
Neutral pH - yellow
Weak base - yellow
Strong base - yellow

Therefore this flower solution changes colour (from yellow through orange to red) as pH decreases (becomes more acidic), but doesn’t change at all as pH increases from neutral to alkaline.
The solution is a good/useful acid pH indicator but useless as a basic/alkali indicator.

#### Dane Red

##### Member
Would the following help with the question in the first attachment?

$\bg_white Pb\left(s\right)+2AgNO_3\left(aq\right)\:\rightarrow \: Pb\left(NO_3\right)_2\left(aq\right)+2Ag\left(s\right)$

$\bg_white n_{Pb}=\frac{20.72\:g}{207.2\:g\:mol^{-1}}=0.1000\:mol$

$\bg_white n_{Ag}=0.100\:L\:\times 1.00\:mol\:L^{-1}=0.100\:mol$

$\bg_white \text{Mole ratio}\: Pb\::\:AgNO_3\:\text{is}\:1\::\:2$

$\bg_white \therefore 0.100\:mol\:AgNO_3\:\text{and}\:0.0500\:mol\: Pb\:\text{used}$

Moles of remaining ions and solids:

$\bg_white Pb^{2+}\left(aq\right)=0.0500\:mol$

$\bg_white Ag^+\left(aq\right)=0\:mol$

$\bg_white NO_3^{\:-}\left(aq\right)=0.100\:mol$

$\bg_white Pb\left(s\right)=0.0500\:mol$

$\bg_white Ag\left(s\right)=0.100\:mol$
What was the formula that you used here

#### Masaken

##### Clown™
What was the formula that you used here
View attachment 37639
that formula is n=cV (number of moles = concentration in mol/L times volume in litres) - this formula is from mod 2

#### Eagle Mum

##### Well-Known Member
What was the formula that you used here
View attachment 37639
It’s the formula to calculate how many moles of silver (as ions in the silver nitrate solution) there are in the initial solution, by multiplying the volume of the solution (0.100L) by the concentration of silver nitrate (1.00 mol/L).

#### Dane Red

##### Member
that formula is n=cV (number of moles = concentration in mol/L times volume in litres) - this formula is from mod 2
How could i forget that, but anyway thanks for clarifying