1Al2(CO3)3 + 6HCL --> 2ALCL3 + 3CO2 + 3H20

NaOH (aq) + HCl(aq) H2O + NaCl (aq)

yuradex [85]

Answer:

The answer to your question is 0.005

Explanation:

Data

Volume of NaOH = 25 ml

[NaOH] = 0.2 M

moles of NaOH = ?

To solve this problem is not necessary to have the chemical reaction. Just use the formula of Molarity and solve it for moles.

Formula

Molarity = moles / volume

-Solve for moles

moles = Molarity x volume

-Convert volume to liters

1000 ml ---------------- 1 l

25 ml ---------------- x

x = (25 x 1) / 1000

x = 0.025 l

-Substitution

moles = 0.2 x 0.025

-Result

moles = 0.005

7 0

3 years ago

An analytical chemist is titrating of a solution of propionic acid with a solution of 224.9 ml of a 0.6100M solution of propioni

Svetllana [295]

Answer: The pH of acid solution is 4.58

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

For KOH:

Molarity of KOH solution = 1.1000 M

Volume of solution = 41.04 mL

Putting values in equation 1, we get:

$1.1000M=\frac{\text{Moles of KOH}\times 1000}{41.04}\\\\\text{Moles of KOH}=\frac{1.1000\times 41.04}{1000}=0.04514mol$

For propanoic acid:

Molarity of propanoic acid solution = 0.6100 M

Volume of solution = 224.9 mL

Putting values in equation 1, we get:

$0.6100M=\frac{\text{Moles of propanoic acid}\times 1000}{224.9}\\\\\text{Moles of propanoic acid}=\frac{0.6100\times 224.9}{1000}=0.1372mol$

The chemical reaction for propanoic acid and KOH follows the equation:

$C_2H_5COOH+KOH\rightarrow C_2H_5COOK+H_2O$

Initial: 0.1372 0.04514

Final: 0.09206 - 0.04514

Total volume of solution = [224.9 + 41.04] mL = 265.94 mL = 0.26594 L (Conversion factor: 1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[\text{salt}]}{[acid]})$

$pH=pK_a+\log(\frac{[C_2H_5COOK]}{[C_2H_5COOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of propanoic acid = 4.89

$[C_2H_5COOK]=\frac{0.04514}{0.26594}$

$[C_2H_5COOH]=\frac{0.09206}{0.26594}$

pH = ?

Putting values in above equation, we get:

$pH=4.89+\log(\frac{(0.04514/0.26594)}{(0.09206/0.26594)})\\\\pH=4.58$

Hence, the pH of acid solution is 4.58

7 0

3 years ago

Consider these chemical equations. N2(g) + 3H2(g) → 2NH3(g) C(s) + 2H2(g) → CH4(g) 4H2(g) + 2C(s) + N2(g) → 2HCN(g) + 3H2(g) Whi

joja [24]

Multiply the second equation by 2

3 0

2 years ago

Read 2 more answers

List out the moles for each element in the following compound ZnBr2

natulia [17]

Answer:

If the colors in a chromatography are able to dissolve and travel up a paper wick, what kind of chemical property do the colors have when mixed with rubbing alcohol?

(You may need to search "Chemical Properties")

4 0

2 years ago

An electrochemical cell has the following standard cell notation.

Alona [7]

Explanation:

The given following standard cell notation.

Mg(s) | Mg^2+ (aq) || Aq^+(aq) | Aq(s)

Oxidation:

$Mg(s)\rightarrow Mg^{2+}+2e^-$ ....(1)

Magnesium metal by loosing 2 electrons is getting converted into magnesium cation. Hence, getting oxidized

Reduction:

$Ag^+(aq)+1e^-\rightarrow Ag(s)$ ...(2)

Silver ion by gaining 1 electrons is getting converted into silver metal. Hence, getting reduced.

Overall redox reaction: (1)+2 × (2)

$Mg(s)+2Ag^+(aq)\rightarrow Mg^{2+}+2Ag(s)$

4 0

3 years ago