Which provides a baseline for an experimental investigation

The law of the conservation of mass states that the total mass before and after a

ehidna [41]

Answer:

<u>reaction are equal.</u>

Explanation:

The law of the conservation of mass states that the total mass before and after a reaction are equal.

In a reaction, the mass of the reactants are always equal to the mass of the products. Nothing ever disappears from the equation, although substances may become gas and "disappear" the air.

6 0

4 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

Increasing the temperature increases the vaporization rate of a liquid because the excess energy is used to break covalent bonds

igor_vitrenko [27]

Answer:

False

Explanation:

False. The molecules of liquid are hold in the liquid state due to intermolecular forces or Van de Waals forces , without affecting the molecule itself and its atomic bonds (covalent bonds). When the temperature increases the kinetic energy of the molecules is higher , therefore they have more possibilities to escape from the attractive intermolecular forces and go to the gas state.

Note however that this is caused because the intermolecular forces are really weak compared to covalent bonds, therefore is easier to break the first one first and go to the gas state before any covalent bond breaks ( if it happens).

A temperature increase can increase vaporisation rate if any reaction is triggered that decomposes the liquid into more volatile compounds , but nevertheless, this effect is generally insignificant compared with the effect that temperature has in vaporisation due to Van der Waals forces.

4 0

3 years ago

draw the lewis structure for CO2, H2CO3, HCO3-, and CO3 2-.Rank these in order of increasing attraction to water molecules. Expl

gavmur [86]

Answer:

The structures are attached in file.

Hydrogen bonding and intermolecular forces is the reason for ranks allotted.

Explanation:

In determining Lewis structure, we calculate the overall number of valence electrons available for bonding. Making carbon (the least electronegative atom) the central atom in the structure, we allocate valence electrons until each atom has achieved stability.

In order of decreasing affinity to water molecules:

$CO_{3}^{2-} > HCO_{3} ^{2-} > H_{2} CO_{3}$

This is due to the fact that the $CO_{3}^{2-}$ will accept protons more readily than the bicarbonate ion, $HCO_{3} ^{2-}$ . Carbonic acid, $H_{2} CO_{3}$ will not accept any more protons, hence it is the least attractive to water molecule, even though soluble.