How do water currents transfer heat different part of the ocean?

Please give explanation as well

jekas [21]

Answer:

Concentration of HCl = $\frac{0.01}{0.134}$ = 0.075L= 75mL

Concentration of KOH = $\frac{0.002}{0.357}$ = 0.0056L = 5.6mL

Concentration of H₂SO₄ = $\frac{0.014}{0.13}$ = 0.108L = 108mL

Explanation:

Procedure

This a problem related to volumetric analysis and we need to find the concentrations of the acids involved in the neurtralization process.

In order to determine the concentrations, we work from the known reactants to the unknown reactants or products.

1. Write the balanced equation of the reactions

2. List the given parameters and work from the known to the unknown. The known is that specie that can give us the number of moles required for this reaction.

3. Check the parameters and make sure that they are in their appropriate units.

4. Obtain the number of moles of the known using the concentration and volume of the reactant using the equation below:

Number of moles = Concentration x Volume.

5. Using the known number of moles, determine that of the unknown by comparing their mole ratios.

6. Since we have obtained the number of moles of the unknown, we can then solve for the concentration of the unknown using the expression below:

$\frac{number of moles}{volume}$

Solution

1. If it takes 67 mL of 0.15 M NaOH to neutralize 134 mL of an HCl solution, what is the concentration of the HCl?

Given parameters:

Volume of base NaOH = 67mL to litres = 67 x 10⁻³ = 0.067L

Concentration of NaOH = 0.15M

Volume of acid HCl = 134mL = 134 x 10⁻³ = 0.134L

Concentration of acid = ?

Equation of reaction: NaOH + HCl → NaCl + H₂O

The known here is the base, NaOH.

Using:

Number of moles of NaOH = Concentration of NaOH x Volume of NaOH

Number of moles of NaOH = 0.15M x 0.067L = 0.01mol

From the equation of the reaction, we know that;

1mole of NaOH reacted with 1mole of HCl

Therefore, 0.01 mole of HCl would also react with 0.01 of NaOH

Now that we know the number of moles HCl, we can now obtain the concentration of HCl required to neutralize NaOH using the equation below:

Concentration of HCl = $\frac{number of moles}{volume}$

Concentration of HCl = $\frac{0.01}{0.134}$ = 0.075L= 75mL

2. If it takes 27.4 mL of 0.050 M H₂SO₄ to neutralize 357 mL of KOH solution, what is the concentration of the KOH solution?

Given parameters:

Concentration of H₂SO₄ = 0.05M

Volume of acid H₂SO₄ = 27.4mL = 27.4 x 10⁻³ = 0.0274L

Volume of base KOH = 357mL to litres = 357 x 10⁻³ = 0.357L

Concentration of KOH = ?

Equation of reaction: 2KOH + H₂SO₄ → K₂SO₄ + 2H₂O

The known here is the acid, H₂SO₄

Using:

Number of moles of H₂SO₄ = Concentration of H₂SO₄ x Volume of H₂SO₄

Number of moles of H₂SO₄ = 0.05M x 0.0274L = 0.001mol

From the equation of the reaction, we know that;

2moles of KOH reacted with 1mole of H₂SO₄

Therefore, 0.001 mole of H₂SO₄ would also react with 0.002 of KOH

Now that we know the number of moles KOH, we can now obtain the concentration of KOH required to neutralize H₂SO₄ using the equation below:

Concentration of KOH = $\frac{number of moles}{volume}$

Concentration of KOH = $\frac{0.002}{0.357}$ = 0.0056L = 5.6mL

3. If it takes 55 mL of 0.5 M NaOH solution to completely neutralize 130 mL of sulfuric acid solution H₂SO₄, what is the concentration of the H₂SO₄ solution?

Given parameters:

Volume of base NaOH = 55mL to litres = 55 x 10⁻³ = 0.055L

Concentration of NaOH = 0.5M

Concentration of H₂SO₄ = ?

Volume of acid H₂SO₄ = 130mL = 130 x 10⁻³ = 0.13L

Equation of reaction: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

The known here is the base, NaOH

Using:

Number of moles of NaOH = Concentration of NaOH x Volume of NaOH

Number of moles of NaOH = 0.5M x 0.055L = 0.028mol

From the equation of the reaction, we know that;

2moles of NaOH reacted with 1mole of H₂SO₄

Therefore, 0.028 mole of NaOH would also react with $\frac{1}{2}$ of 0.028, 0.014mole of H₂SO₄

Now that we know the number of moles H₂SO₄, we can now obtain the concentration of H₂SO₄ required to neutralize NaOH using the equation below:

Concentration of H₂SO₄ = $\frac{number of moles}{volume}$

Concentration of H₂SO₄ = $\frac{0.014}{0.13}$ = 0.108L = 108mL

5 0

4 years ago

Name and write the formula of the ion that makes a solution acidic

BARSIC [14]

Name and write the formula of the ion that makes a solution acidic

H hydrogen

7 0

4 years ago

Which of the following is the correct rank order of the diffusion rates of either Cl2 or H2S through air at a pressure of 1.0 at

Radda [10]

Answer:

Cl2 at 250 K < Cl2 at 300 K < H2S at 250 K < H2S at 300 K.

Explanation:

3 0

2 years ago

What is the difference between polyatomic an diatomic?

motikmotik

Monatomic chemical elements contain only one atom.

Polyatomic chemical compounds have two or more atoms.

3 0

4 years ago

Read 2 more answers

Phosphoric acid is a triprotic acid ( K a1 = 6.9 × 10 − 3 Ka1=6.9×10−3, K a2 = 6.2 × 10 − 8 Ka2=6.2×10−8, and K a3 = 4.8 × 10 −

PilotLPTM [1.2K]

<u>Answer:</u> To calculate the pH of the buffer composed of $H_2PO_4^-\text{ and }HPO_4^{-2}$ , we use the $K_a2$

<u>Explanation:</u>

Phosphoric acid is a triprotic acid and it will undergo three dissociation reaction each having their respective dissociation constants.

The chemical equation for the first dissociation reaction follows:

$H_3PO_4\rightleftharpoons H_2PO_4^-+H^+;K_a1=6.9\times 10^{-3}$

The chemical equation for the second dissociation reaction follows:

$H_2PO_4^-\rightleftharpoons HPO_4^{2-}+H^+;K_a2=6.2\times 10^{-8}$

The chemical equation for the third dissociation reaction follows:

$HPO_4^{2-}\rightleftharpoons PO_4^{3-}+H^+;K_a3=4.8\times 10^{-13}$

To form a buffer composed of $H_2PO_4^-\text{ and }HPO_4^{-2}$ , we use the $K_a$ of second dissociation process

To calculate the $pK_a$ , we use the equation:

$pK_a=-\log (K_a)\\\\pK_a=-\log(6.2\times 10^{-8})\\\\pK_a=7.21$

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

$pH=pK_a2+\log(\frac{[\text{conjugate base}]}{[\text{weak acid}]})$

$pH=pK_a2+\log(\frac{[HPO_4^{2-}]}{[H_2PO_4^-]})$

We are given:

$pK_a2$ = negative logarithm of second acid dissociation constant of phosphoric acid = 7.21

$[HPO_4^{2-}]$ = concentration of conjugate base

$[H_2PO_4^{-}]$ = concentration of weak acid

Hence, to calculate the pH of the buffer composed of $H_2PO_4^-\text{ and }HPO_4^{-2}$ , we use the $K_a2$

8 0

3 years ago