When a metal and____ bond, they do not share electrons.

Starting with the unbalanced equation for the combustion of ethanol, C2H5OH(l)+O2(g)→CO2(g)+?H2O(l) what coefficient should be p

lianna [129]

Answer:

C2H5OH(l) + O2(g) → CO2(g) + ?H2O(l)

The coefficient should be 3

Explanation:

C2H5OH(l) + O2(g) → CO2(g) + ?H2O(l)

The chemical equation is a reaction between ethanol and oxygen . The reaction yield is carbon dioxide and water. Balancing the chemical equation requires one to make the number of atoms on the reactant side be equal to the number of atoms on the product side.

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

The hydrogen atom on the left hand side of the equation is 6. Therefore, the number of atom of hydrogen should be 6 on the right hand side. The number of oxygen atom on the left hand side is 7. Therefore, the number of Oxygen atom on the right hand side should 7 . Oxygen atom in carbon dioxide(right hand side) is 4 atoms, therefore, it requires 3 atoms of oxygen to make the oxygen atoms 7.

Using 3 as the coefficient of water in the equation will make hydrogen 6 and oxygen 3 .Adding the 3 oxygen to the already 4 atoms of oxygen in carbon dioxide will make oxygen 7 atoms

6 0

3 years ago

A closed, frictionless piston-cylinder contains a gas mixture with the following composition on a mass basis: 40% carbon dioxide

Hitman42 [59]

Answer:

W=-37.6kJ, therefore, work is done on the system.

Explanation:

Hello,

In this case, the first step is to compute the moles of each gas present in the given mixture, by using the total mixture weight the mass compositions and their molar masses:

$n_{CO_2}=0.8kg*0.4*\frac{1kmolCO_2}{44kgCO_2}= 0.00727kmolCO_2\\\\n_{O_2}=0.8kg*0.25*\frac{1kmolO_2}{32kgO_2}=0.00625kmolO_2\\ \\n_{Ne}=0.8kg*0.35*\frac{1kmolNe}{20.2kgNe}=0.0139kmolNe$

Next, the total moles:

$n_T=0.00727kmol+0.00625kmol+0.0139kmol=0.02742kmol$

After that, since the process is isobaric, we can compute the work as:

$W=P(V_2-V_1)$

Therefore, we need to compute both the initial and final volumes which are at 260 °C and 95 °C respectively for the same moles and pressure (isobaric closed system)

$V_1=\frac{n_TRT_1}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(260+273)K}{450kPa}=0.27m^3\\ \\V_2=\frac{n_TRT_2}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(95+273)K}{450kPa}=0.19m^3$

Thereby, the magnitude and direction of work turn out:

$W=450kPa(0.19m^3-0.27m^3)\\\\W=-37.6kJ$

Thus, we conclude that since it is negative, work is done on the system (first law of thermodynamics).

Regards.

7 0

3 years ago

In an experiment you are asked to react hydrochloric acid with sodium hydroxide. When measuring the volume of the reactants, whi

Bingel [31]

What are the options? based on the volume it can differ but in general pipettes and graduated cylinder are the best.

7 0

3 years ago

Which of the following most likely happens when the volume of a gas increases?

mafiozo [28]

the pressure of the gas increases

6 0

3 years ago

I need help please!!! Name the parts making up the female part of the flower and state what each part does. Please use your own

Greeley [361]

<h2>Answer:-</h2>

The female part is the pistil. The pistil usually is located in the center of the flower and is made up of three parts: the stigma, style, and ovary. The stigma is the sticky knob at the top of the pistil. It is attached to the long, tubelike structure called the style.

6 0

2 years ago