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inysia [295]
3 years ago
11

Ainda sobre a combustao completa do metanol que pode ser representada pela equação nao-balanceada abaixo. Quando se utilizam 8,0

mols de metanol nessa reação, qual é a quantidade de materia (mol) de agua produzida (em mols)?
(olhe a imagem por favor)

Chemistry
1 answer:
Radda [10]3 years ago
3 0

Answer:

n_{H_2O}=16molH_2O

Explanation:

Hello,

In this case, given the properly balanced reaction:

CH_3OH+\frac{3}{2} O_2\rightarrow CO_2+2H_2O

If 8.0 moles of methanol react, we can compute the produced grams of water by noticing the 1:2 molar ratio between them in the chemical reaction (stoichiometric coefficients):

n_{H_2O}=8molCH_3OH*\frac{2molH_2O}{1molCH_3OH} \\\\n_{H_2O}=16molH_2O

Best regards.

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You have 500.0 ml of a buffer solution containing 0.30 m acetic acid (ch3cooh) and 0.20 m sodium acetate (ch3coona). what will t
Nataly [62]
First, we should get moles acetic acid = molarity * volume

                                                                =0.3 M * 0.5 L

                                                                =  0.15 mol

then, we should get moles acetate = molarity * volume

                                                           = 0.2 M * 0.5L

                                                           = 0.1 mol

then, we have to get moles of OH- which added:

moles OH- = molarity  * volume

                   = 1 M    * 0.02L

                  = 0.02 mol

when the reaction equation is:


                 CH3COOH  +  OH-  → CH3COO-   +  H2O


moles acetic acid after adding OH- = (0.15-0.02) 
                                             
                                                            =  0.13M                                       

moles acetate after adding OH- =  (0.1 + 0.02)

                                                      =   0.12 M

Total volume = 0.5 L + 0.02 L= 0.52 L

∴[acetic acid] = moles acetic acid after adding OH- / total volume

                        = 0.13mol / 0.52L

                       = 0.25 M

and [acetate ) = 0.12 mol / 0.52L
 
                        = 0.23 M

by using H-H equation we can get PH:

PH = Pka + ㏒[salt/acid]

when we have Ka = 1.8 x 10^-5

∴Pka = -㏒Ka 

        = -㏒ 1.8 x 10^-5

       = 4.7

So by substitution:

∴ PH = 4.7 + ㏒[acetate/acetic acid]

         = 4.7 + ㏒(0.23/0.25)

        = 4.66
6 0
3 years ago
Classify each process according to direction of heat flow.
MrRissso [65]

Answer:

heat flows in

heat flows out

heat flows out

Explanation:

6 0
4 years ago
Read 2 more answers
There are four cis,trans isomers for 2-isopropyl-5-methylcyclohexanol, where the cis,trans designations of the substituents are
andreev551 [17]

The image is attached with the answer.

<h3>What are Isomers ?</h3>

Isomers are compounds which have same empirical formula , same number of atoms are present but they have difference in arrangement if the atoms.

It is given that

There are four cis, trans isomers for 2-isopropyl-5-methylcyclohexanol

the cis, trans designations of the substituents are made relative to the oh group

The image is attached with this answer.

To know more about Isomers

brainly.com/question/13422357

#SPJ1

5 0
2 years ago
A balloon has a pressure of 3.1 atm at a volume of 155 ml. if the temperature is held constant, what is the volume (ml) if the p
GarryVolchara [31]

Answer:

45.8 mL

Explanation:

If all variables are held constant, the new volume can be found using the Boyle's Law equation. The equation looks like this:

P₁V₁ = P₂V₂

In this equation, "P₁" and "V₁" represent the initial pressure and volume. "P₂" and "V₂" represent the final pressure and volume. You can find the new volume by plugging the given values into the equation and simplifying.

P₁ = 3.1 atm                        P₂ = 10.5 atm

V₁ = 155 mL                       V₂ = ? mL

P₁V₁ = P₂V₂                                                 <----- Boyle's Law equation

(3.1 atm)(155 mL) = (10.5 atm)V₂                <----- Insert values

480.5 = (10.5 atm)V₂                                  <----- Multiply 3.1 and 155

45.8 = V₂                                                    <----- Divide both sides by 10.5

4 0
2 years ago
If 180 grams of potassium iodide is dissolved in 100 cm3 of water at 30oC, a(n) _______________ solution is formed.
olga nikolaevna [1]

Super saturated solution is formed.

<u>Explanation:</u>

Solubility is the property of any substance's capacity, that is the solute of the substance is dissolved in the given solvent to form the solution. We have three different types of solution, unsaturated, saturated and supersaturated solution.

  • Unsaturated solution is a solution with lesser amount of solute than its solubility at equilibrium.
  • Saturated solution is a solution with the maximum solute dissolved in the solvent.
  • Super saturated solution is a solution with more solute than it is required.

The solubility of KI at 30°C is 153 g / 100 ml. Here 180 g of KI in 100 ml of water at 30°C is given, which has more solute than required, so it is super saturated solution.

6 0
3 years ago
Read 2 more answers
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