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2 years ago

5

PLS HELP STOICHIOMETRY/CHEM!!!

1 answer:

Mariulka [41]2 years ago

7 0

The molecular formula of the compound that we are required to find is the compound C4H8O8

<h3>What is empirical formula?</h3>

The empirical formula of a compound is a formula that shows the ratio of each atom present in the compound. We will start by dividing each mass with the relative atomic mass of the atom.

Carbon - 48.38 g/12 Hydrogen - 6.74 g/1 Oxygen - 53.5 g/16

Carbon - 4 Hydrogen - 6.74 Oxygen - 8.9

Dividing through by the lowest ratio;

Carbon - 4/4 Hydrogen - 6.74/4 Oxygen 8.9/4

Carbon 1 Hydrogen 2 Oxygen 2

The empirical formula is CH2O2.

To obtain the molecular formula; brainly.com/question/11588623

[12 + 2 + 32]n = 180

n = 180/[12 + 2 + 32]

n =4

The compound C4H8O8

Learn more about empirical formula:

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What is the molecular weight of bromoform if 2.50X10^-2 mol weighs 6.33 g

babymother [125]

Jwwnsnsnxjxjdnsmzmmxzmzm

8 0

3 years ago

When CO2 decomposes into oxygen and carbon, it gives a gram ratio of 2.67:1 O2:C. When a 32.4g of CO2 decomposes, how many grams

Rufina [12.5K]

Answer : The mass of carbon and oxygen produced is 8.83 g and 23.6 g respectively.

Explanation :

Law of conservation of mass : It states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.

This also means that total mass on the reactant side must be equal to the total mass on the product side.

The balanced chemical reaction will be,

$CO_2\rightarrow O_2+C$

As we are given:

$\text{Mass of }O_2}:\text{Mass of }C=2.67:1{$

According to the law of conservation of mass,

Total mass of $CO_2$ = Mass of $O_2$ + Mass of C

Total mass of $CO_2$ = 2.67 + 1 = 3.67 g

Now we have to calculate the mass of $O_2$ and C.

$\text{Mass of }O_2=\frac{\text{Given mass of }CO_2}{\text{Total mass of }CO_2}\times \text{Given mass of }O_2=\frac{32.4g}{3.67g}\times 2.67=23.6g$

and,

$\text{Mass of }C=\frac{\text{Given mass of }CO_2}{\text{Total mass of }CO_2}\times \text{Given mass of }C=\frac{32.4g}{3.67g}\times 1=8.83g$

Therefore, the mass of carbon and oxygen produced is 8.83 g and 23.6 g respectively.

6 0

3 years ago

A 13.5g sample of gold is heated, then placed in a calorimeter containing 60g of water. The temperature of water increases from

sweet-ann [11.9K]

Answer:

$T_i~=163.1$ ºC

Explanation:

We have to start with the variables of the problem:

Mass of water = 60 g

Mass of gold = 13.5 g

Initial temperature of water= 19 ºC

Final temperature of water= 20 ºC

<u>Initial temperature of gold= Unknow</u>

Final temperature of gold= 20 ºC

Specific heat of gold = 0.13J/gºC

Specific heat of water = 4.186 J/g°C

Now if we remember the <u>heat equation</u>:

$Q_H_2_O=m_H_2_O*Cp_H_2_O*deltaT$

$Q_A_u=m_A_u*Cp_A_u*deltaT$

We can relate these equations if we take into account that <u>all heat of gold is transfer to the water</u>, so:

$m_H_2_O*Cp_H_2_O*deltaT=~-~m_A_u*Cp_A_u*deltaT$

Now we can <u>put the values into the equation</u>:

$60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C=-(13.5~g*0.13~J/g{\circ}C*(20-T_i)~{\circ}C)$

Now we can <u>solve for the initial temperature of gold</u>, so:

$T_i~=(\frac{60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C}{13.5~g*0.13~J/g{\circ}C})+20$

$T_i~=163.1$ ºC

I hope it helps!

5 0

3 years ago

Is bleach liquid starch? <br> Yes or No

Jobisdone [24]

Answer:

O it's not

Explanation:

Have a great day!

8 0

2 years ago

What is 0.3m equal to mm?

madreJ [45]

0.3×100cm=300cm

300cm×10mm=3000

4 0

2 years ago