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3 years ago

A solution with a higher concentration of hydroxide ions (oh-) than hydrogen ions (h+)

1 answer:

5 0

A solution with a higher concentration of hydroxide ions than hydrogen ions is basic solution.

This solution formed by Base dissolved in water and release hydroxide ions.

The PH of the solution is greater than 7

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A 2.832 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 6.439 grams of CO2 and 2

strojnjashka [21]

Answer: The empirical and molecular formula for the given organic compound is $C_2H_2O_4$ and $C_6H_{12}O_2$ respectively.

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=6.439g$

Mass of $H_2O=2.636g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 6.439 g of carbon dioxide, $\frac{12}{44}\times 6.439=1.756g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 2.636 g of water, $\frac{2}{18}\times 2.636=0.293g$ of hydrogen will be contained.

Mass of oxygen in the compound = (2.832) - (1.756 + 0.293) = 0.783 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.756g}{12g/mole}=0.146moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.293g}{1g/mole}=0.293moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.783g}{16g/mole}=0.049moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.049 moles.

For Carbon = $\frac{0.146}{0.049}=2.97\approx 3$

For Hydrogen = $\frac{0.293}{0.049}=5.97\approx 6$

For Oxygen = $\frac{0.049}{0.049}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 6 : 1

Hence, the empirical formula for the given compound is $C_3H_{6}O_1=C_3H_6O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 116.2 g/mol

Mass of empirical formula = 58 g/mol

Putting values in above equation, we get:

$n=\frac{116.2g/mol}{58g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(3\times 2)}H_{(6\times 2)}O_{(1\times 2)}=C_6H_{12}O_2$

Thus, the empirical and molecular formula for the given organic compound is $C_3H_6O$ and $C_6H_{12}O_2$ respectively.

7 0

3 years ago

Difference in solubility products and molar solubility?

klasskru [66]

I dont know kid, gotta ask someone else.

3 0

3 years ago

By mistake, you added salt instead of sugar to the oil. How can you remove the salt?

Ghella [55]

Explanation:

To remove the salt from the oil, I will add water to dissolve the salt from it.

Oil is an organic molecule that is non-polar

Salt is polar ionic compound

Salt will not dissolve in the oil.

Take the mixture.
Add water to it.
Water and oil are immiscible
Shake the new heterogeneous mixture vigorously.
leave to settle.
Oil will come on top of the water.
You can skim off the oil layer on top.
Then heat the water and salt solution.
This leaves the oil behind.

learn more:

Mixture brainly.com/question/1832352

#learnwithBrainly

5 0

3 years ago

Read 2 more answers

A 73.6 g sample of aluminum is heated to 95.0°C and dropped into 100.0 g of water at 20.0°C. If the resulting temperature of the

Softa [21]

Answer:

The specific heat of aluminium is 0.875 J/g°C

Explanation:

Step 1: Data given

The mass of the aluminium sample = 73.6 grams

Initial temperature of the sample = 95.0 °C

Mass of water = 100.0 grams

Initial temperature of water = 20.0 °C

Final temperature of water and aluminium = 30.0 °C

The specific heat of water = 4.184 J/g°C

Step 2: Calculate the specific heat of aluminium

Q gained = Q lost

Qwater = -Qaluminium

Q = m*c*ΔT

m(aluminium) * c(aluminium) * ΔT(aluminium) = - m(water) * c(water) * ΔT(aluminium)

⇒ mass of aluminium = 73.6 grams

⇒ c(aluminium) = TO BE DETERMINED

⇒ ΔT(aluminium) = The change of temperature = T2 - T1 = 30 .0 °C - 95.0 °C = -65.0°C

⇒ mass of water = 100.0 grams

⇒ c(water ) = The specific heat of water = 4.184 J/g°C

⇒ ΔT(water) = The change of temperature of water = T2 - T1 = 30.0 - 20.0 = 10.0 °C

73.6g * c(aluminium) * -65.0 °C = 100.0g * 4.184 J/g°C * 10.0°C

-4784 * c(aluminium) = -4184

c(aluminium) = 0.875 J /g°C

The specific heat of aluminium is 0.875 J/g°C

7 0

3 years ago

Which of the following is an example of an endothermic process? (2 points) Gasoline burning in a combustion engine A chocolate b

frozen [14]

A chocolate bar melting in a hot car

8 0

3 years ago

Read 2 more answers