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3 years ago

An aqueous solution of potassium sulfate (K2SO4) has a freezing point of -2.24

1 answer:

7 0

An aqueous solution of potassium sulfate exhibits colligative properties. Colligative properties are properties that depends on the concentration of a substance in a solution. These properties are freezing point depression, vapor pressure lowering, osmotic pressure and boiling point elevation. For this problem we use the concept of freezing point depression since we are given the freezing point of the solution. Freezing point depression is as:

ΔT = -k(f) x m x i
-2.24 - 0 = -1.86 x m x 3
m = 0.4014

Thus, the molality of the solution is 0.4014.

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3 years ago

Glucose (c6h12o6) has a single carbonyl group (-c=o) in its linear form. based on the number of oxygen atoms in glucose, how man

sveta [45]

Glucose is a hexose monosaccharide. It is one of the three major monosaccharides along with fructose and galactose. These are carbohydrates with a general formula of Cₓ(H₂O)ₓ, where x could be any number.

Now, you don't have to know the structural formula of the glucose to answer this. Just account all the elements in the glucose. You know that there are 6 oxygen atoms all in all. One of them belongs to the single carbonyl group. Consequently, that would mean that the remaining 5 oxygen atoms bond with hydrogen atoms to form 5 OH groups.

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3 years ago

Need help ASAP with this question

Karo-lina-s [1.5K]

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Explanation: zoom on on the red section and look at the key for what red means.

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3 years ago

How many grams do 6.534e+24 molecules of phosphoric acid weigh?

fredd [130]

Answer:

1,063 grams H₃PO₄

Explanation:

To find the mass of phosphoric acid (H₃PO₄), you should (1) convert molecules to moles (via Avogadro's number) and then (2) convert moles to grams (via molar mass from periodic table).

Molar Mass (H₃PO₄): 3(1.008 g/mol) + 30.974 g/mol + 4(15.998 g/mol)

Molar Mas (H₃PO₄): 97.99 g/mol

6.534 x 10²⁴ molecules H₃PO₄ 1 mole 97.99 g
--------------------------------------------- x ------------------------------------- x --------------
6.022 x 10²³ molecules 1 mole

= 1,063 grams H₃PO₄

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2 years ago

The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. Combustion of

gavmur [86]

Answer: The empirical formula for the given compound is $C_3H_6O$

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:

Conversion factor: 1 g = 1000 mg

Mass of $CO_2=6.32mg=0.00632g$

Mass of $H_2O=2.58g=0.00258g$

Mass of compound = 2.78 mg = 0.00278 g

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 0.00632 g of carbon dioxide, $\frac{12}{44}\times 0.00632=0.00172g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 0.00258 g of water, $\frac{2}{18}\times 0.00258=0.000286g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.00278) - (0.00172 + 0.000286) = 0.000774 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.00172g}{12g/mole}=1.43\times 10^{-4}moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.000286g}{1g/mole}=2.86\times 10^{-4}moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.000774g}{16g/mole}=4.83\times 10^{-5}moles$

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 $4.83\times 10^{-5}mol$

For Carbon = $\frac{1.43\times 10^{-4}}{4.83\times 10^{-5}}=2.96\approx 3$

For Hydrogen = $\frac{2.86\times 10^{-4}}{4.83\times 10^{-5}}=5.92\approx 6$

For Oxygen = $\frac{4.83\times 10^{-5}}{4.83\times 10^{-5}}=1$

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 $C_3H_{6}O_1=C_3H_6O$

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3 years ago