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4 years ago

Calculating [H.]

Chemistry

1 answer:

klemol [59]4 years ago

5 0

Answer:

The answer to your question is given below below:

Explanation:

1. Data obtained from the question:

Concentration of Hydroxide ion, [OH-] = 1x10^-1 M

Concentration of Hydrogen ion, [H+] =..?

The concentration of Hydrogen ion and hydroxide ion are related with the following formula:

[H+] x [OH-] = 1x10^-14

[H+] x 1x10^-1 = 1x10^-14

Divide both side by 1x10^-1

[H+] = 1x10^-14 / 1x10^-1

[H+] = 1x10^-13M

2. Data obtained from the question include:

pH = 2

Hydrogen ion concentration, [H+] =..?

pH = - log [H+]

2 = - log [H+]

-2 = log [H+]

[H+] = antilog (-2)

[H+] = 0.01M

3. Data obtained from the question:

Concentration of Hydroxide ion, [OH-] = 1x10^-8 M

Concentration of Hydrogen ion, [H+] =..?

The concentration of Hydrogen ion and hydroxide ion are related with the following formula:

[H+] x [OH-] = 1x10^-14

[H+] x 1x10^-8 = 1x10^-14

Divide both side by 1x10^-8

[H+] = 1x10^-14 / 1x10^-8

[H+] = 1x10^-6M

4. Data obtained from the question include:

pOH = 9

Hydrogen ion concentration, [H+] =..?

First we shall determine the pH.

The pH and pOH are related with the following formula:

pH + pOH = 14

pH + 9 = 14

Collect like terms

pH = 14 - 9

pH = 5

Now, we can obtain the [H+] as follow:

pH = - log [H+]

pH = 5

5 = - log [H+]

-5 = log [H+]

[H+] = antilog (-5)

[H+] = 1x10^-5M

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Which choice best describes the specific heat of water compared with that of metals?

n200080 [17]

Answer: C. The specific heat of water is greater than the specific heat of metals.

The specific heat is defined as the amount required to raise the temperature of a unit mass of a substance by 1 degree Celsius.

This is expressed mathematically as

Q= mc∆T

Where Q is the energy/heat required which is measured in Joules.

m is the mass (grams)

c is the specific heat which is measured in joule/gram degree Celsius.

∆T- change in temperature

Substance which has a high specific heat require a lot of heat for its temperature to be raised by one degree. On the other hand substances with lower specific heat require only little amount of heat for its temperature to be raised by one degree.

Consider an equal mass of metal and water. If both are heated at the same time,the metal would become hotter than the water much faster. This is because the specific heat of the metal is lower than the water. Hence it requires only a little heat for its temperature to raised by one degree.

Thus we can conclude that the specific heat of water is much greater than that of a metal.

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3 years ago

The Mystery Salt

ryzh [129]

Question 1 :

To determine whether the salt is KCl or KNO3, one should look for the difference between them in terms of their physicochemical properties, such as their solubility.

Since we have the solubilité of KCl and KNO3, we can use the property of solubility to determine if the mystery salt is KCl or KNO3.

Question 2:

We will try to reproduce the conditions to determine the solubility of the salt at 37°C.

We will put into the beaker 100ml of water (equivalent to 100g) and dissolve a defined quantity of the salt (the number should be between the solubility of the KCl (37g) and KNO3 (30g) so between 30g and 37g).

Let's dissolve for example 32g of the salt, then, heat with the hotplate until the temperature of the beaker content will be 35 ° C (use the thermometer to determine the exact temperature).

Why?

This manipulation aims to determine the solubility of our mystery salt to know if it is KNO3 or KCl. In our conditions, we will obtain two different possibilities depending on if the salt is KCl or KNO3, this justifies why we took a quantity between 30g and 37g of salt.

If it is KNO3 (solubility of 30g/ml) we will observe a precipitation in the beaker because we exceed its solubility.

If it is KCl (37g/100) we will not observe a precipitate since we did not attempt the solubility of KCl

Question 3:

Finally to determine the composition of salt: we know that the solubility of KCL is 37g / 100ml (that is to say if we dissolve a higher mass (38g for example), we will observe a precipitation of salt) and that the solubility of KNO3 is 30g / 100ml (that is to say if we dissolve an upper mass (32g for example), we will observe a precipitation of salt)

In our experiment, 32g of salts were dissolved. If it is KCl, we will not observe a precipitate since the minimum concentration to start having a precipitate is not yet reached (37g / 100ml).

If it is KNO3, a precipitate will be observed since the minimum concentration to start having a precipitate is not yet reached (30g / 100ml).

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3 years ago

What is flame emission spectroscopy

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Flame emission is a method of chemical that uses intensity of light from a flame.

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3 years ago

Calculate the mass of silver chloride in grams required to make 5.8 g of silver plating.

Fudgin [204]

Silver chloride is simply AgCl while silver is Ag so we see that there is one mole to one mole correspondence. First calculate the moles of Ag:

moles Ag = 5.8 g / ( 107.87 g/mol)

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Then the mass of AgCl required:

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4 years ago

You have 26.7 mL of 0.061 mol/L aqueous potassium hydroxide (KOH(aq)) in a conical flask. In a burette

Natasha2012 [34]

Answer:

9.47 mL

Explanation:

The reaction that takes place is:

2KOH + H₂SO₄ → K₂SO₄ + 2H₂O

First we calculate how many KOH moles reacted, using the given concentration and volume of KOH solution:

0.061 mol/L = 0.061 mmol/mL

0.061 mmol/mL * 26.7 mL = 1.6287 mmol KOH

Then we convert KOH moles into H₂SO₄ moles, using the stoichiometric coefficients:

1.6287 mmol KOH * $\frac{1mmolH_2SO_4}{2mmolKOH}$ = 0.8144 mmol H₂SO₄

Finally we calculate the required volume of the H₂SO₄ solution, using the number of moles and given concentration:

0.8144 mmol ÷ 0.086 mmol/mL = 9.47 mL

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3 years ago