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2 years ago

A solution has a [H+] of 0.20 M. Find the pH of this solution.

Chemistry

1 answer:

Rus_ich [418]2 years ago

4 0

The pH of this solution : 0.699

<h3>Further explanation </h3>

pH is the degree of acidity of a solution that depends on the concentration of H⁺ ions. The greater the value the more acidic the solution and the smaller the pH.

pH = - log [H⁺]

So that the two quantities between pH and [H⁺] are inversely proportional because they are associated with negative values.

A solution whose value is different by n has a difference in the concentration of H⁺ ion of 10ⁿ.

A solution has a [H+] of 0.20 M, so [H⁺]=0.2

$\tt pH=-log[H^+]\\\\pH=-log[0.2]\\\\pH=1-log~2\\\\pH=0.699$

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Calculate how many grams of the product form when 16.7 g of calcium metal completely reacts. Assume that there is more than enou

swat32

39.96 g product form when 16.7 g of calcium metal completely reacts.

<h3>What is the stoichiometric process?</h3>

Stoichiometry is a section of chemistry that involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data.

Equation:

$Ca(s) + Cl_2(g)$ → $CaCl_2(s)$

In this case, for the undergoing reaction, we can compute the grams of the formed calcium chloride by noticing the 1:1 molar ratio between calcium and it (stoichiometric coefficients) and using their molar mass of 40 g/mol and 111 g/mol by using the following stoichiometric process:

$m_{ca_C_l_2}$ = 16.7 g Ca x $\frac{1 mol \;of \;Ca}{40g Ca}$ x $\frac{1 mol \;of \;CaCl_2}{1 mol \;Ca}$ x $\frac{111g of \;CaCl_2}{1 mol \;CaCl_2}$

$m_{ca_C_l_2}$ = 39.96 g

Hence, 39.96 g product form when 16.7 g of calcium metal completely reacts.

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2 years ago

Commercial grade fuming nitric acid contains about 90.0% HNO3 by mass with a density of 1.50 g/mL, calculate the molarity of the

andrew-mc [135]

Answer:

Since molarity is defined as moles of solute per liter of solution, we need to find the number of moles of nitric acid, and the volume of solution.

molar mass of nitric acid (HNO3) = 1 + 14 + (3x16) = 15 + 48 = 63 g/mole

1.50 g/ml x 1000 ml = 1500 g/liter

1500 g/liter x 0.90 = 1350 g/liter of pure HNO3 (the 0.9 is to correct for the fact that it is 90% pure)

1350 g/liter x 1 mole/63 g = 21.43 moles/liter = 21 Molar HNO3

= 21 Molar of HNO3

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2 years ago

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Exercise 2:

vagabundo [1.1K]

Answer:

1. KCLO3------>KCL + 3/2O2(g)

2. 122.5g/mol

3. 0.2mol

4. 18.5g

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2 years ago

!!HELP PLS!! A compound sample contains 0.783g of C, 0.196 g of H, 0.521 g 0 and the molecular formula molar mass is 184.27g/mol

Delvig [45]

Answer:

First

divide each element by its Molecular Mass to get their respective moles

Then Divide through by the lowest of the moles

You'll have the ratio of Carbon Hydrogen and Oxygen to be

C2H3O

Given Molecular Mass=184.27

C2H3On=184.27

n(12x2 + 1x3 + 16) =184.27

Evaluating this... You'll have n=4.3

Pls check if you assigned the correct value to each element

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2 years ago

1. What is the molarity of a solution prepared by dissolving 3.11 grams of NaOH in

Elena L [17]

The molarity of a solution prepared by dissolving 3.11 grams of NaOH in

enough water to make 300 milliliters of solution is 0. 256 mol/ L

Explanation:

<h3>What is Molarity?</h3>

Molarity (M) is the amount of a substance in a given volume of solution. It is otherwise known as the number of moles of solute in a certain amount of solution.

The unit of molarity is mol/L

<h3>Formula for calculating molarity :</h3>

Molarity = number of moles of solute divided by the volume of the solution\

<h3>Parameters </h3>

Molarity = ?

Number of moles is calculated using :

Mass of the solute = 3.11 grams of NaOH

Molar mass of the solute = Na + O + H

Let's input the atomic numbers of the elements: Na (23) , O (16) , H (1)

Molar mass of NaOH = 23 + 16 + 1 = 40 g/ mol

Number of moles = 3.11 ÷ 40 = 0.077 mol

Volume of the solution is measured in Liters

Let's convert 300 milliters to liters = 300 divided by 1000 = 0.3 Liters

Molarity = 0.077 ÷ 0.3 = 0. 256 mol/ L

Therefore, the Molarity of the solution is 0. 256 mol/ L

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1 year ago