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3 years ago

How to write a balanced chemical equation for the standard formation reaction of liquid octanol?

1 answer:

3 0

To write a balanced chemical equation for the standard formation reaction of liquid octanol form the octanol from its elements — C, H, and O.

The standard enthalpy of formation of a compound is the change in enthalpy when 1 mol of the compound is formed from its constituent elements. All substances must be in their most stable states at 100 kPa and 25 °C.

write a thermochemical equation for the formation of 1 mol of octanol.

8C(s) + 9H₂(g) + ½O₂(g) → C₈H₁₇OH(l)

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a laboratory procedure calls for making 510.0 mL of a 1.6 M KNO3 solution. How much KNO3 in grams is needed

barxatty [35]

Answer:

82.416 g of KNO ₃ is needed to produce 510.0 mL of a 1.6 M KNO ₃ solution.

Explanation:

Since molarity is the number of moles of solute that are dissolved in a given volume, calculated by dividing the moles of solute by the volume of the solution, the following rule of three can be applied: if in 1 L (1,000 mL) of KNO₃ there are 1.6 moles of the compound present, in 510 mL how many moles will there be?

$moles=\frac{510 mL*1.6 moles}{1000 mL}$

moles= 0.816

Being the molar mass of the elements:

K: 39 g/mole
N: 14 g/mole
O: 16 g/mole

So the molar mass of the compound KNO₃ is:

KNO₃= 39 g/mole + 14 g/mole + 3*16 g/mole= 101 g/mole

Now I can apply the following rule of three: if in 1 mole of KNO₃ there are 101 g, in 0.816 moles how much mass is there?

$mass=\frac{0.816 moles*101 grams}{1 mole}$

mass= 82.416 grams

82.416 g of KNO ₃ is needed to produce 510.0 mL of a 1.6 M KNO ₃ solution.

4 0

2 years ago

Whats the answer to this

Natasha2012 [34]

I think the answer you have selected is right, I'm not sure tho

3 0

3 years ago

One canned juice drink is 15% orange juice; another is 5% orange juice. How many liters of each should be mixed together in orde

Maksim231197 [3]

It uses elimination againLet A be 15% juice and B is 5% juice
A+B = 100.15A + 0.05B = 0.11*10 = 1.1Multiply 2nd equation by 100 to get rid of decimals
A+B = 1015A + 5B = 110

7 0

3 years ago

Cinnamon owes its flavor and odor to cinnamaldehyde (C9H8O). Determine the boiling point elevation of a solution of 92.7 mg of c

Flauer [41]

Answer:

The boiling point elevation is 3.53 °C

Explanation:

∆Tb = Kb × m

∆Tb is the boiling point elevation of the solution

Kb is the molal boiling point elevation constant of CCl4 = 5.03 °C/m

m is the molality of the solution is given by moles of solute (C9H8O) divided by mass of solvent (CCl4) in kilogram

Moles of solute = mass/MW =

mass = 92.7 mg = 92.7/1000 = 0.0927 g

MW = 132 g/mol

Moles of solute = 0.0927/132 = 7.02×10^-4 mol

Mass of solvent = 1 g = 1/1000 = 0.001 kg

m = 7.02×10^-4 mol ÷ 0.001 kg = 0.702 mol/kg

∆Tb = 5.03 × 0.702 = 3.53 °C (to 2 decimal places)

6 0

3 years ago

How many grams of sodium acetate are in solution in the third beaker?

Kipish [7]

Answer:

46g of sodium acetate.

Explanation:

The data is: Precipitation from a supersaturated sodium acetate solution. The solution on the left was formed by dissolving 156g of the salt in 100 mL of water at 100°C and then slowly cooling it to 20°C. Because the solubility of sodium acetate in water at 20°C is 46g per 100mL of water, the solution is supersaturated. Addition of a sodium acetate crystal causes the excess solute to crystallize from solution.

The third solution is the result of the equilibrium in the solution at 20°C. As the maximum quantity that water can dissolve of sodium acetate at this temperature is 46g per 100mL and the solution has 100mL there are 46g of sodium acetate in solution. The other sodium acetate precipitate because of decreasing of temperature.

I hope it helps!

6 0

3 years ago