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2 years ago

Calcula el pH de las siguientes sustancias formadas durante una erupción volcánica:

Chemistry

1 answer:

Virty [35]2 years ago

8 0

Calculate the pH of the following substances formed during a volcanic eruption:

• Acid rain if the [H +] is 1.9 x 10-5

• Sulfurous acid if [H +] = 0.10

• Nitric acid if [H +] = 0.11

<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⁺]

pH acid rain

$\tt pH=-log[1.9\times 10^{-5}]\\\\pH=5-log1.9\\\\pH=4.72$

pH Sulfurous acid

$\tt pH=-log[10^{-1}]\\\\pH=1$

pH Nitric acid

$\tt pH=-log[11\times 10^{-2}]\\\\pH=2-log~11=0.959$

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3 years ago

A sulfuric acid solution containing 571.3 g of h2so4 per liter of aqueous solution has a density of 1.329 g/cm3. Part a calculat

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Mass percentage of a solution is the amount of solute present in 100 g of the solution.

Given data:

Mass of solute H2SO4 = 571.3 g

Volume of the solution = 1 lit = 1000 ml

Density of solution = 1.329 g/cm3 = 1.329 g/ml

Calculations:

Mass of the given volume of solution = 1.329 g * 1000 ml/1 ml = 1329 g

Therefore we have:

571.3 g of H2SO4 in 1329 g of the solution

Hence, the amount of H2SO4 in 100 g of solution= 571.3 *100/1329 = 42.987

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3 years ago

The independent variable has control and affects the

Stella [2.4K]

Answer:

The independent variable is the condition that you change in an experiment. It is the variable you control.

Explanation:

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3 years ago

Read 2 more answers

In Lab 10 you make a stock solution of salicylic acid, and then four dilutions. The stock solution is made by diluting 5.00 ml o

Zigmanuir [339]

Answer:

Stock solution = 1.25 *10^-3 M

Dilution 1 = 5*10^-4 M

Dilution 2= 3.75 * 10^-4 M

Dilution 3 = 2.5 *10^-4 M

Dilution 4 = 1.25 *10^-4 M

Explanation:

Step 1: Data given

The stock solution is made by diluting 5.00 ml of 1.250 x 10-2 M salicylic acid in 50.00 mL of solution.

Step 2: Calculate the concentration of the stock solution:

M1*V1 = M2*V2

⇒ with M1 = the initial concentration = 1.250 *10^-2 M

⇒ with V1 = 5 mL = 5*10^-3 L

⇒ with M2 = TO BE DETERMINED

⇒ with V2 = 50 mL = 50 *10^-3 L

M2 = (M1*V1)/V2

M2 = (1.250 *10^-2 * 5*10^-3 L) / 50 *10^-3

M2 = 0.00125 M = 1.25 *10^-3 M

Step 3: Calculate dilution 1

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 10 *10^-3)/(25*10^-3L)

M2 = 0.0005 M = 5*10^-4 M

Step 4: Calculate dilution 2

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 10 * 7.5*10^-3)/(25*10^-3)

M2 = 0.000375 M = 3.75 * 10^-4 M

Step 5: Calculate dilution 3

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 5*10^-3) /(25*10^-3)

M2 = 0.00025 M = 2.5 *10^-4 M

Step 6: Calculate dilution 4

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 2.5*10^-3)/(25*10^-3)

M2 = 0.000125 M = 1.25 *10^-4 M

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3 years ago

A certain compound was found to have the molecular formula C5H12O2. To which of the following compound classes could the compoun

IgorC [24]

Answer : The given compound belongs to ether and alcohol.

Explanation :

The chemical formula of the given compound is, $C_5H_{12}O_2$

First we have to calculate the degree of unsaturation.

Formula used:

Degree of unsaturation = $\frac{2C+2+N-X-H}{2}$

where,

C = number of carbon

H = number of hydrogen

N = number of nitrogen

X = number of halogen

Degree of unsaturation = $\frac{2\times 5+2+0-0-12}{2}=0$

The degree of unsaturation is, 0 that means there is no double or triple bond in the compound only single bond is present between the atoms.

Thus, the given compound belongs to ether and alcohol.

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3 years ago