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2 years ago

The pH of a solution can be calculated using which formula?

Chemistry

1 answer:

VikaD [51]2 years ago

5 0

Answer:

The formula for calculating pH is pH=−log[H_3O+ ]

pH is the negative logarithm (to base 10) of hydronium ion concentration

The pH Formula can also be expressed as

PH= - log[H+ ]

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The depletion of ozone (O3) in the stratosphere has been a matter of great concern among scientists in recent years. It is belie

Deffense [45]

Answer: Nitric oxide is the limiting reagent. The number of moles of excess reagent left is 0.0039 moles. The amount of nitrogen dioxide produced will be 0.7912 g.

Explanation:

To calculate the number of moles, we use the equation

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

For ozone:

Given mass of ozone = 0.827 g

Molar mass of ozone = 48 g/mol

Putting values in above equation, we get:

$\text{Moles of ozone}=\frac{0.827g}{48g/mol}=0.0172mol$

For nitric oxide:

Given mass of nitric oxide = 0.635 g

Molar mass of nitric oxide = 30.01 g/mol

Putting values in above equation, we get:

$\text{Moles of nitric oxide}=\frac{0.635g}{30.01g/mol}=0.0211mol$

For the given chemical equation:

$O_3+NO\rightarrow O_2+NO_2$

By Stoichiometry of the reaction:

1 mole of ozone reacts with 1 mole of nitric oxide.

So, 0.0172 moles of ozone will react with = $\frac{1}{1}\times 0.0172=0.0172moles$ of nitric oxide

As, given amount of nitric oxide is more than the required amount. So, it is considered as an excess reagent.

Thus, ozone is considered as a limiting reagent because it limits the formation of product.

Amount of excess reagent (nitric oxide) left = 0.0211 - 0.0172 = 0.0039 moles

By Stoichiometry of the reaction:

1 mole of ozone produces 1 mole of nitrogen dioxide.

So, 0.0172 moles of ozone will react with = $\frac{1}{1}\times 0.0172=0.0172moles$ of nitrogen dioxide

Now, calculating the mass of nitrogen dioxide from equation 1, we get:

Molar mass of nitrogen dioxide = 46 g/mol

Moles of nitrogen dioxide = 0.0172 moles

Putting values in equation 1, we get:

$0.0172mol=\frac{\text{Mass of nitrogen dioxide}}{46g/mol}\\\\\text{Mass of nitrogen dioxide}=0.7912g$

Hence, nitric oxide is the limiting reagent. The number of moles of excess reagent left is 0.0039 moles. The amount of nitrogen dioxide produced will be 0.7912 g.

8 0

3 years ago

What type of reaction is most likely to occur when barium reacts with fluorine? combustion synthesis decomposition single replac

Lana71 [14]

Answer is: synthesis.
Chemical reaction: Ba + F₂ → BaF₂.
Synthesis is type of reaction where two or more compounds (in this reaction barium and fluorine) react to form one product (in this reaction BaF₂).
BaF₂ - barium fluoride, salt, white cubic crystals, soluble in methanol and ethanol.

8 0

3 years ago

Read 2 more answers

If you are using 3.00% (mass/mass) hydrogen peroxide solution and you determine that the mass of solution required to reach the

Maurinko [17]

Answer:

0.004522 moles of hydrogen peroxide molecules are present.

Explanation:

Mass by mass percentage of hydrogen peroxide solution = w/w% = 3%

Mass of the solution , m= 5.125 g

Mass of the hydrogen peroxide = x

$w/w\% = \frac{x}{m}\times 100$

$3\%=\frac{x}{5.125 g}\times 100$

$x=\frac{3\times 5.125 g}{100}=0.15375 g$

Mass of hydregn pervade in the solution = 0.15375 g

Moles of hydregn pervade in the solution :

$=\fraC{ 0.15375 g}{34 g/mol}=0.004522 mol$

0.004522 moles of hydrogen peroxide molecules are present.

5 0

2 years ago

PLEASE HELP ME ASAP URGENT IM STRESSED AND STUCK

Genrish500 [490]

Is there any more info lol
But from thinking I got A=DH^2*BC

8 0

2 years ago

Ethanol, C2H6O, is most often blended with gasoline - usually as a 10 percent mix - to create a fuel called gasohol. Ethanol is

weeeeeb [17]

Answer:

This means 463 grams of ethanol would provide less amount of energy

Explanation:

Step 1: Data given

Heat of combustion of ethanol = 326.7 kcal/mol

The heat of combustion of octane = 1.308*10³ kcal/mol

Mass of octane = 463 grams

Molar mass octane = 114.23 g/mol

Molar mass ethanol = 46.07 g/mol

Step 2: Calculate moles octane

Moles octane = mass octane / molar mass octane

Moles octane = 463 grams / 114.23 g/mol

Moles octane = 4.05 moles

Step 3: Calculate energy of combustion of 4.05 moles octane

Combustion of 1 mol octane gives us: 1.308 * 10³ kcal/mol

Combustion of 4.05 moles octane gives us 4.05 * 1.308 * 10³ kcal/mol = 5.30 * 10³ kcal

This means the combustion reaction of 463 grams of octane gives us 5.30 * 10³ kcal

Step 4:

Heat of combustion of ethanol = 326.7 kcal/mol

OR in words: combustion of 1 mol ethanol gives us 326.7 kcal energy

Moles ethanol = 463 grams / 46.07 g/mol

Moles ethanol = 10.05 moles

Since combustion of 1 mol ethanol gives us 326.7 kcal

10.05 moles ethanol will give us = 10.05 * 326.7 = 3283.3 kcal = 3.28 * 10³ kcal

5.30 * 10³ kcal > 3.28 * 10³ kcal

This means 463 grams of ethanol would provide less amount of energy

3 0

3 years ago