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3 years ago

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In the simulation, open the Custom mode. The beaker will be filled to the 0.50 L mark with a neutral solution. Set the pH to 11.

30 by using the green arrows adjacent to the pH value indicated on the probe in the solution. Once you adjust the pH, note the corresponding OH− ion concentration in M as given in the graphic on the left side of the simulation. Make sure to select the option "Concentration (mol/L)" above the graphic. Find the pOH of the solution. Express your answer numerically to two decimal places.

1 answer:

Mila [183]3 years ago

4 0

Answer:

Hydroxyl ion concentration is $5.08\times10^{-11}M$ .

Explanation:

From the given,

pH of the solution = 11.30

pH+pOH = 14

11.30 + pOH = 14

pOH = 14-11.30 = 2.7

The pOH of the solution is 2.7

<u>Concentration of hydroxyl ion:</u>

$2.7= -log[OH^{-}]$

$log[OH^{-}]= -2.7=5.08\times10^{-11}M$

Therefore, hydroxylion concentration is $5.08\times10^{-11}M$ .

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What does iron do for the human body(two paragraphs)

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Answer:

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Explanation:

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6 0

3 years ago

Calculate the number of moles contained in 550ml of carbon dionide at stp

g100num [7]

The answer is 20 I think

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3 years ago

Hydrogen peroxide decomposes to water and oxygen at constant pressure by the following reaction: 2H2O2(l) → 2H2O(l) + O2(g) ΔH =

Mashcka [7]

<u>Answer:</u> The amount of heat released is -7.203 kJ

<u>Explanation:</u>

The given chemical equation follows:

$2H_2O_2(l)\rightarrow 2H_2O(l )+O_2(g);\Delta H=-196kJ$

To calculate the enthalpy change for 1 mole of the hydrogen peroxide, we use unitary method:

When 2 moles of hydrogen peroxide is reacted, the enthalpy of the reaction is -196 kJ

So, when 1 mole of hydrogen peroxide will react, the enthalpy of the reaction will be $\frac{-196}{2}\times 1=-98kJ$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of hydrogen peroxide = 2.50 g

Molar mass of hydrogen peroxide = 34 g/mol

Putting values in above equation, we get:

$\text{Moles of hydrogen peroxide}=\frac{2.50g}{34g/mol}=0.0735mol$

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

$\Delta H_{rxn}=\frac{q}{n}$

where,

$q$ = amount of heat released

n = number of moles = 0.0735 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction = -98 kJ/mol

Putting values in above equation, we get:

$-98kJ/mol=\frac{q}{0.0735mol}\\\\q=(-98kJ/mol\times 0.0735mol)=-7.203kJ$

Hence, the amount of heat released is -7.203 kJ

8 0

3 years ago

Complete combustion of 8.60 g of a hydrocarbon produced 26.5 g of CO2 and 12.2 g of H2O. What is the empirical formula for the h

amid [387]

The empirical formula of the hydrocarbon is $C_2H_3$ if combustion of 8.60 g of a hydrocarbon produced 26.5 g of $CO_2$ and 12.2 g of $H_2O$ .

<h3>What is an empirical formula?</h3>

A chemical formula showing the simplest ratio of elements in a compound rather than the total number of atoms in the molecule $CH_2O$ is the empirical formula for glucose.

1 mole of carbon dioxide contains a mass of 44 g, out of which 12 g are carbon.

Hence, in this case the mass of carbon in 8.46 g of $CO_2$ :

( $\frac{12}{44}$ ) × 8.46 = 2.3073 g

1 mole of water contains 18 g, out of which 2 g is hydrogen;

Therefore, 2.6 g of water contains;

( $\frac{2}{18}$ × 2.6 = 0.2889 g of hydrogen.

Therefore, with the amount of carbon and hydrogen from the hydrocarbon, we can calculate the empirical formula.

We first calculate the number of moles of each,

Carbon = $\frac{2.3073}{12}$ = 0.1923 moles

Hydrogen = $\frac{0.2889}{1}$ = 0.2889 moles

Then, we calculate the ratio of Carbon to hydrogen by dividing by the smallest number value;

Carbon : Hydrogen

$\frac{0.1923}{0.1923}$ : $\frac{0.2889}{0.1923}$

1 : 1.5

(1 : 1.5) 2

= 2 : 3

Hence, the empirical formula of the hydrocarbon is $C_2H_3$ .

Learn more about the empirical formula here:

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2 years ago

The amount of energy in the bonds of the reactants plus any energy absorbed equals the amount of energy in the bonds of the prod

Vinvika [58]

The amount of energy in the bonds of the reactants plus any energy absorbed equals the amount of energy in the bonds of the products plus any energy <span>given off when the products are formed.

This statement explains and verifies the law of conservation of energy during the chemical reactions. This law states that energy can neither be created nor destroyed, but can change from one form to another.</span>

4 0

3 years ago

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