How the heck do i do this

Given 8.25 g of butanoic acid and excess ethanol, how many grams of ethyl butyrate would be synthesized, assuming a complete 100

mash [69]

Answer:

10.87 g of Ethyl Butyrate

Solution:

The Balance Chemical Equation is as follow,

H₃C-CH₂-CH₂-COOH + H₃C-CH₂-OH → H₃C-CH₂-CH₂-COO-CH₂-CH₃ + H₂O

According to equation,

88.11 g (1 mol) Butanoic Acid produces = 116.16 g (1 mol) Ethyl Butyrate

So,

8.25 g Butanoic Acid will produce = X g of Ethyl Butyrate

Solving for X,

X = (8.25 g × 116.16 g) ÷ 88.11 g

X = 10.87 g of Ethyl Butyrate

8 0

2 years ago

What is the speed at which molecules or atoms move dependent on temperature and state of matter.

den301095 [7]

The hotter it gets, the faster molecules move, solid form is in low temperature, liquid in medium temperature and gas in high temperature.

3 0

3 years ago

A graduated cylinder was filled to 20.3 mL with water. A solid object weighing 73.05 g was immersed in the water, raising the me

victus00 [196]

The density of the solid object will be 2.63 g/mL

<h3>What is density?</h3>

Density of objects = mass/volume.

Recall that an object will always displace its own volume when placed in a liquid.

Volume of the solid object = Cylinder reading after immersing the object in the water - cylinder reading before immersing the object in the water.

= 48.1 - 20.4

= 27.8 mL

Mass of the solid object = 73.05 g

Density of the object = 73.05/27.8

= 2.63 g/mL

More on density can be found here: brainly.com/question/15164682

#SPJ1

3 0

8 months ago

What is the boiling point of a solution of 76 g of water dissolved in 500 mL of acetic acid, CH3COOH?

tatuchka [14]

Answer:

127.3° C, (This is not a choice)

Explanation:

This is about the colligative property of boiling point.

ΔT = Kb . m . i

Where:

ΔT = T° boling of solution - T° boiling of pure solvent

Kb = Boiling constant

m = molal (mol/kg)

i = Van't Hoff factor (number of particles dissolved in solution)

Water is not a ionic compound, but we assume that i = 2

H₂O → H⁺ + OH⁻

T° boling of solution - 118.1°C = 0.52°C . m . 2

Mass of solvent = Solvent volume / Solvent density

Mass of solvent = 500 mL / 1.049g/mL → 476.6 g

Mol of water are mass / molar mass

76 g / 18g/m = 4.22 moles

These moles are in 476.6 g

Mol / kg = molal → 4.22 m / 0.4766 kg = 8.85 m

T° boling of solution = 0.52°C . 8.85 m . 2 + 118.1°C = 127.3°C

6 0

3 years ago

Read 2 more answers

When a 0.245-g sample of benzoic acid is combusted in a bomb calorimeter, the temperature rises 1.643 ∘C . When a 0.260-g sample

sveticcg [70]

Answer:

The heat of combustion per moles of caffeine is 4220 kJ/mol

Explanation:

Step 1: Data given

⇒ When benzoic acid sample of 0.245 grams is burned the temperature rise is 1.643 °C

⇒ When 0.260 gram of caffeine is burned, the temperature rise is 1.436 °C

⇒ Heat of combustion of benzoic acid = 26.38 kJ/g

<u>Step 2:</u> Calculate the heat released: for combustion of benzoic acid

0.245 g benzoic acid * 26.38 kJ/g = 6.4631 kJ

<u>Step 3</u>: Calculate the heat capacity of the calorimeter:

c = Q/ΔT

Q = 6.4631 kJ / 1.643°C = 3.934 kJ/ °C

<u>Step 4:</u> Calculate moles of a 0.260 g sample of caffeine:

Moles caffeine = Mass caffeine / Molar mass caffeine

0.260 grams/ 194.19 g/mol = 0.0013389 moles

Step 5: Calculate heat released: for combustion of caffeine

Q = c * ΔT

Q = 3.934 kJ/°C * 1.436 °C = 5.65 kJ

Step 6: Calculate the heat of combustion per mole of caffeine

5.65 kJ / 0.0013389 moles = 4219.9 kJ/mol ≈ 4220 kJ/mol

The heat of combustion per moles of caffeine is 4220 kJ/mol

4 0

3 years ago