I’m stuck on number 2

A 0.8715 g sample of sorbic acid, a compound first obtained from the berries of a certain ash tree, is burned completely in oxyg

aivan3 [116]

<h3>Answer;</h3>

C3H4O

<h3>Explanation;</h3>

Empirical formula is the simplest formula of a compound;

Molar mass CO2 = 44.01

Mass of CO2 produced = 2.053 g

Mass of carbon in original sample = 12.01/44.01 × 2.053

= 0.5603g

Molar mass H2O = 18

Mass of H in original sample = 2/18 ×0.5601

= 0.0622 g

Thus; original sample contained 0.5603g C and 0.0622g H. The balance of the sample was O

Mass of O = 0.8715 - (0.5603 + 0.0622) = 0.249g

The mole ratio of C:H:O will be;

Moles C = 0.5603/12 = 0.0467

Moles H = 0.0622

Moles O = 0.249/16 = 0.01556

C:H:O = 0.0467:0.0622:0.01556

Divide through by 0.01556:

C:H:O = 3:4:1

Empirical formula is thus C3H4O

8 0

4 years ago

Read 2 more answers

Please help this is due today which ones are reactive and which arent?

BARSIC [14]

Answer:

I' really don't know I'm sorry

4 0

3 years ago

Calculate the gravitional force in between two object of mass 25kg and 20kg if distance between them is 5m?

astraxan [27]

Answer:

Explanation:The equation for this is F = -GmM/R^2 where the minus sign says the force is attractive m is 10 kg, M is 20 kg and R is 5 meters. If you crunch the numbers you get an answer of:

7 0

3 years ago

Commercial hydrochloric acid (HCl) is typically labeled as being 38.0 % (weight %). The density of HCl is 1.19 g/mL. a) What is

Strike441 [17]

Answer:

For a: The molarity of commercial HCl solution is 12.39 M.

For b: The molality of commercial HCl solution is 16.79 m.

For c: The volume of commercial HCl solution needed is 2.42 L.

Explanation:

We are given:

Mass % of commercial HCl solution = 38 %

This means that 38 grams of HCl is present in 100 grams of solution.

To calculate the volume of solution, we use the equation:

$Density=\frac{Mass}{Volume}$

Density of HCl solution = 1.19 g/mL

Mass of solution = 100 g

Putting values in above equation:

$1.19g/mL=\frac{100g}{\text{Volume of solution}}\\\\\text{Volume of solution}=84.034mL$

For a:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Molarity of solution = ?

Molar mass of HCl = 36.5 g/mol

Volume of solution = 84.034 mL

Mass of HCl = 38 g

Putting values in above equation, we get:

$\text{Molality of commercial HCl solution}=\frac{38\times 1000}{36.5\times 84.034}\\\\\text{Molality of commercial HCl solution}=12.39M$

Hence, the molarity of commercial HCl solution is 12.39 M.

For b:

To calculate the molality of solution, we use the equation:

$Molarity=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

Where,

$m_{solute}$ = Given mass of solute (HCl) = 38 g

$M_{solute}$ = Molar mass of solute (HCl) = 36.5 g/mol

$W_{solvent}$ = Mass of solvent = 100 - 38 = 62 g

Putting values in above equation, we get:

$\text{Molality of commercial HCl solution}=\frac{38\times 1000}{36.5\times 62}\\\\\text{Molality of commercial HCl solution}=16.79m$

Hence, the molality of commercial HCl solution is 16.79 m.

For c:

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted solution

We are given:

$M_1=6M\\V_1=5.00L\\M_2=12.39M\\V_2=?L$

Putting values in above equation, we get:

$6\times 5=12.39\times V_2\\\\V_2=2.42L$

Hence, the volume of commercial HCl solution needed is 2.42 L.

5 0

3 years ago

The purpose of a scientific investigation is

torisob [31]

Is to find the problem to the investigation

5 0

4 years ago