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2 years ago

What is the pH of a 0.75 M HNO3 solution

Chemistry

2 answers:

pogonyaev2 years ago

6 0

Answer: The pH of the solution is 0.125

Explanation:

pH is the negative logarithm of hydrogen ion concentration present in a solution.

To calculate the pH of the reaction, we use the equation:

$pH=-\log[H^+]$

The chemical equation for the dissociation of nitric acid follows:

$HNO_3\rightarrow H^++NO_3^-$

By stoichiometry of the reaction:

1 mole of nitric acid produces 1 mole of hydrogen ions and 1 mole of nitrate ions.

So,

$[H^+]=0.75M$

Putting values in above equation, we get:

$pH=-\log(0.75)\\\\pH=0.125$

Hence, the pH of the solution is 0.125

Sauron [17]2 years ago

4 0

Hello!

The dissociation reaction of HNO₃ is the following:

HNO₃ → H⁺ + NO₃⁻

This is a strong acid, so the concentration of HNO₃ would be the same as the concentration of H⁺. The formula for pH is the following:

$pH=-log([H_3O^{+}])=-log(0,75M)=0,12$

So, the pH would be 0,12

Have a nice day!

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If you compared 1 m solutions, was a 1 m nacl solution more or less hypertonic than a 1 m sucrose solution? what is your evidenc

igor_vitrenko [27]

When you say the solution is hypertonic, it means that the solution has a higher osmotic pressure. The formula for this is:

P = iMRT,
for strong electrolytes, i = number of ions.
for nonelectrolytes, i = 1

1. The P for sucrose solution which is a nonelectrolyte (assuming room temp):
P = (1)(1m)(8.314 J/mol-K)(298 K)
P = 2477.572 Pa

The P for NaCl solution, which is a strong electrolyte:
P = (2)(1 m)(8.314)(298 K)
P = 4955.144 Pa

So, that means that NaCl is more hypertonic than the sucrose solution.

2. For the second question, the P for the combination of 1 m glucose (nonelectrolyte) and 1 m sucrose is:
P = (1)(1 m)(8.314)(298 K) + (1)(1)(8.314)(298 K) = 4955.144 Pa
In this case, the osmotic pressures are now equal. It is not hypertonic, but isotonic.

7 0

2 years ago

If the lead can be extracted with 92.5% efficiency, what mass of ore is required to make a lead sphere with a 3.50 cm radius? le

ElenaW [278]

The volume of sphere can be calculated using the following formula:

$V=\frac{4}{3}\pi r^{3}$

Here, r is radius of the sphere which is 3.5 cm. Putting the value,

$V=\frac{4}{3}\pi r^{3}=\frac{4}{3}(3.14)(3.5cm)^{3}=180 cm^{3}$

This is equal to the volume of lead, density of lead is $11.34 g/cm^{3}$ thus, mass of lead can be calculated as follows:

m=d×V

Putting the values,

$m=11.34 g/cm^{3}\times 180 cm^{3}=2041.2 g$

Let the mass of ore be 1 g, 68.6% of galena is obtained by mass, thus, mass of galena obtained will be 0.686 g.

Now, 86.6% of lead is obtained from this gram of galena, thus, mass of lead will be:

m=0.686×0.866=0.5940 g

Therefore, 0.5940 g of lead is obtained from 1 g of ore for 100% efficiency, thus, for 92.5% efficiency

$m=\frac{92.5}{100}\times 0.5940=0.54945 g$

1 g of lead obtain from $\frac{1}{0.54945}=1.82$ grams of ore.

Thus, 2041.2 g of lead obtain from:

$2041.2\times 1.82=3.715\times 10^{3}g or 3.715 kg$

Therefore, mass of ore required to make lead sphere is 3.715 kg.

6 0

2 years ago

Copper(II) sulfide, CuS, is used in the development of aniline black dye in textile printing. What is the maximum mass of CuS wh

Naya [18.7K]

Answer:

1.82 g is the maximum mass of CuS.

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For $CuCl_2$ :

Molarity = 0.500 M

Volume = 38.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 38.0×10⁻³ L

Thus, moles of $CuCl_2$ :

$Moles=0.500 \times {38.0\times 10^{-3}}\ moles$

Moles of $CuCl_2$ = 0.019 moles

For $(NH_4)_2S$ :

Molarity = 0.600 M

Volume = 42.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 42.0×10⁻³ L

Thus, moles of $(NH_4)_2S$ :

$Moles=0.600 \times {42.0\times 10^{-3}}\ moles$

Moles of $(NH_4)_2S$ = 0.0252 moles

According to the given reaction:

$CuCl_2_{(aq)}+(NH_4)_2S_{(aq)}\rightarrow CuS_{(s)}+2NH_4Cl_{(aq)}$

1 mole of $CuCl_2$ reacts with 1 mole of $(NH_4)_2S$

So,

0.019 mole of $CuCl_2$ reacts with 0.019 mole of $(NH_4)_2S$

Moles of $(NH_4)_2S$ = 0.019 mole

Available moles of $(NH_4)_2S$ = 0.0252 mole

Limiting reagent is the one which is present in small amount. Thus, $CuCl_2$ is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of $CuCl_2$ gives 1 mole of $CuS$

0.019 mole of $CuCl_2$ gives 0.019 mole of $CuS$

Moles of $CuS$ formed = 0.019 moles

Molar mass of $CuS$ = 95.611 g/mol

Mass of $CuS$ = Moles × Molar mass = 0.019 × 95.611 g = 1.82 g

1.82 g is the maximum mass of CuS.

5 0

2 years ago

A sample of fluorine gas is confined in a 5.0-L container at 0.432 atm and 37 °C. How many moles of gas are in this sample?

Ahat [919]

Answer:

About 0.08486 moles

Explanation:

PV=nRT, when P is the pressure in atmospheres, V is the volume in liters, n is the number of moles, R is the universal gas constant, and T is the temperature in Kelvin.

$0.432 \cdot 5= n \cdot 0.0821 \cdot 310$