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3 years ago

Hydrocarbons that contain a double or triple bond between carbons are considered unsaturated because

Chemistry

1 answer:

ipn [44]3 years ago

8 0

Answer:

alkenes and alkynes are considered as unsaturated hydrocarbons because they are capable of reacting with hydrogen in the presence of a catalyst to form a saturated hydrocarbon (alkanes)

Explanation:

A saturated compound contains a chain of carbon atoms joined by single bonds, with hydrogen atoms attached to each carbon atom.

Propane for example, CH₃-CH₂-CH₃.

It is saturated because all the carbon atoms holds maximum number of hydrogen atoms.

Alkenes such as Propene (CH₂=CH-CH₃) and alkynes like Propyne,

(CH₃-C≡C-H) are unsaturated because the central carbon atoms contain less number of hydrogen atoms than they possibly could.

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a) Calculatethe molality, m, of an aqueous solution of 1.22 M sucrose, C12H22O11. The density of the solution is 1.12 g/mL.b) Wh

Contact [7]

Answer:

a) 1,74 molal

b) 37,2 %

c) 0,03

Explanation:

We are going to define sucrose as solute, water as solvent and the mix of both, the solution.

Let´s start with the data:

$Molarity = M = \frac{1,22 mol solute}{lts solution}$

We can assume as a calculus base, 1 liter of solution. So, in 1 liter of solution we have 1,22 moles of solute:

$1 lts solution * \frac{1,22 moles solute}{lts solution}=1,22 moles solute$

Knowing that the molality (m) is defined as mol of solute/kgs solvent, we have to calculate the mass of solvent on the solution. Remember our calculus base (1 lts of solution). In 1 lts of solution we have 1120 grams of solution.

$1 lts solution * \frac{1,12 grs solution}{mL solution}*\frac{1000 mL solution}{1 lts solution} = 1120 grs of solution$

With the molecular weight of solute (Sum of: for carbon = 12*12=144; for hydrogen = 1*22=22 and for oxygen = 16*11=176. Final result = 342 grs per mol), we can obtain the mass of solute:

$1,22 mol solute*\frac{342 grs solute}{1 mol solute} = 417,24 grs solute$

Now, the mass of solvent is: mass solvent = mass of solution - mass of solute. So, we have: 1120 - 417,24 = 702,76 grs of solvent = 0,70276 Kgs of solvent

$molality = m = \frac{1,22 mol solute}{0,70276 kgs solvent}= 1,74 molal$

For b) question we have that the mass percent of solute is hte ratio between the mass of solute and the mass of solution. So,

$%(w/w) = \frac{417,24 grs solute}{1120 grs solution} = 37,2%$

For c) question we have that the mole fraction of solute is the ratio between moles of solute and moles of solution. Let's calculate the moles of solution as follows: Moles solution = moles solute + moles solvent. First we have that the moles of solvent are (remember that the molecular weight of water for this calculus is 18 grs per mol):

$702,76 grs solvent*\frac{1 mol solvent}{18 grs solvent} = 39,04 moles solvent$

So, we have the moles of solution: 1,22 moles of solute + 39,04 moles of solvent = 40,26 moles of solution

Finally, we have:

$Mol frac solute = \frac{1,22 mol solute}{40,26 mol solution}= 0,03$

6 0

3 years ago

What kind of bond is formed when lithium and fluorine combine to form lithium fluoride?

cupoosta [38]

Answer:

ionic bond

Explanation:

4 0

3 years ago

Read 2 more answers

How many kilojoules of energy are required to heat 50.0 g of ethyl alcohol from 60.0 °C to 78.4 °C and vaporize it? The specific

Leno4ka [110]

Answer:

44.62 kJ

Explanation:

Firstly, we calculate the energy needed to heat the liquid (ethyl alcohol) by using the formula:

Q = m × c × ∆T

Where;

Q = Amount of heat (J)

m = mass (g)

c = specific heat of ethyl alcohol = 2.138 J/g°C

∆T = change in temperature (°C)

According to the information given in this question;

Q = ?, m = 50.0g, ∆T = (78.4°C - 60°C) = 18.4°C

Therefore, using Q = mc∆T

Q = 50 × 2.138 × 18.4

Q (amount of energy needed to heat ethyl alcohol) = 1966.96 J

Next, we calculate and add the amount of heat needed to vaporize by using the formula;

How many kilojoules of energy are required to heat 50.0 g of ethyl alcohol from 60.0 °C to 78.4 °C and vaporize it? The specific heat of ethyl alcohol is heat of vaporization is 853 J/g.

3 0

3 years ago

During studies of the reaction below,

Leni [432]

Answer: The percent yield of the nitrogen gas is 11.53 %.

Explanation:

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For NO:

Given mass of NO = 11.5 g

Molar mass of NO = 30 g/mol

Putting values in equation 1, we get:

$\text{Moles of NO}=\frac{11.5g}{30g/mol}=0.383mol$

For $N_2O_4$ :

Given mass of $N_2O_4$ = 102.1 g

Molar mass of $N_2O_4$ = 92 g/mol

Putting values in equation 1, we get:

$\text{Moles of }N_2O_4=\frac{102.1g}{92g/mol}=1.11mol$

For the given chemical reactions:

$2N_2H_4(l)+N_2O_4(l)\rightarrow 3N_2(g)+4H_2O(g)$ ......(2)

$N_2H_4(l)+2N_2O_4(l)\rightarrow 6NO(g)+2H_2O(g)$ .......(3)

Calculating the experimental yield of nitrogen gas:

By Stoichiometry of the reaction 3:

6 moles of NO is produced from 2 moles of $N_2O_4$

So, 0.383 moles of NO will be produced from = $\frac{2}{6}\times 0.383=0.128mol$ of $N_2O_4$

By Stoichiometry of the reaction 2:

1 mole of $N_2O_4$ produces 3 moles of nitrogen gas

So, 0.128 moles of $N_2O_4$ will produce = $\frac{3}{1}\times 0.128=0.384mol$ of nitrogen gas

Now, calculating the experimental yield of nitrogen gas by using equation 1, we get:

Moles of nitrogen gas = 0.384 moles

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

$0.384mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=(0.384mol\times 28g/mol)=10.75g$

Calculating the theoretical yield of nitrogen gas:

By Stoichiometry of the reaction 2:

1 mole of $N_2O_4$ produces 3 moles of nitrogen gas

So, 1.11 moles of $N_2O_4$ will produce = $\frac{3}{1}\times 1.11=3.33mol$ of nitrogen gas

Now, calculating the theoretical yield of nitrogen gas by using equation 1, we get:

Moles of nitrogen gas = 3.33 moles

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

$3.33mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=(3.33mol\times 28g/mol)=93.24g$

To calculate the percentage yield of nitrogen gas, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of nitrogen gas = 10.75 g

Theoretical yield of nitrogen gas = 93.24 g

Putting values in above equation, we get:

$\%\text{ yield of nitrogen gas}=\frac{10.75g}{93.24g}\times 100\\\\\% \text{yield of nitrogen gas}=11.53\%$

Hence, the percent yield of the nitrogen gas is 11.53 %.

7 0

3 years ago

(3 points)

Paul [167]

An increased number of rod cells

5 0

4 years ago

Read 2 more answers