The law of conservation of mass or principle of mass conservation states that for any system closed to all transfers of matter and energy, the mass of the system must remain constant over time, as system's mass cannot change, so quantity cannot be added nor removed. Hence, the quantity of mass is conserved over time.
The law implies that mass can neither be created nor destroyed, although it may be rearranged in space, or the entities associated with it may be changed in form. For example, in chemical reactions, the mass of the chemical components before the reaction is equal to the mass of the components after the reaction. Thus, during any chemical reaction and low-energy thermodynamic processes in an isolated system, the total mass of the reactants, or starting materials, must be equal to the mass of the products.
The concept of mass conservation is widely used in many fields such as chemistry, mechanics, and fluid dynamics. Historically, mass conservation was demonstrated in chemical reactions independently by Mikhail Lomonosov and later rediscovered by Antoine Lavoisier in the late 18th century. The formulation of this law was of crucial importance in the progress from alchemyto the modern natural science of chemistry.
The conservation of mass only holds approximately and is considered part of a series of assumptions coming from classical mechanics. The law has to be modified to comply with the laws of quantum mechanics and special relativityunder the principle of mass-energy equivalence, which states that energy and mass form one conserved quantity. For very energetic systems the conservation of mass-only is shown not to hold, as is the case in nuclear reactions and particle-antiparticle annihilation in particle physics.
Mass is also not generally conserved in open systems. Such is the case when various forms of energy and matter are allowed into, or out of, the system. However, unless radioactivity or nuclear reactions are involved, the amount of energy escaping (or entering) such systems as heat, mechanical work, or electromagnetic radiation is usually too small to be measured as a decrease (or increase) in the mass of the system.
For systems where large gravitational fields are involved, general relativity has to be taken into account, where mass-energy conservation becomes a more complex concept, subject to different definitions, and neither mass nor energy is as strictly and simply conserved as is the case in special relativity.
Answer:
2.56 grams of H₂S is needed to produce 18.00g of PbS if the H2S is reacted with an excess (unlimited) supply of Pb(CH₃COO)₂
Explanation:
The balanced reaction is:
Pb(CH₃COO)₂ + H₂S → 2 CH₃COOH + PbS
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) they react and produce:
- Pb(CH₃COO)₂: 1 mole
- H₂S: 1 mole
- CH₃COOH: 2 moles
- PbS: 1 mole
In this case, to know how many grams of H₂S are needed to produce 18.00 g of PbS, it is first necessary to know the molar mass of the compounds H₂S and PbS and then to know how much it reacts by stoichiometry. Being:
- H: 1 g/mole
- S: 32 g/mole
- Pb: 207 g/mole
The molar mass of the compounds are:
- H₂S: 2* 1 g/mole + 32 g/mole= 34 g/mole
- PbS: 207 g/mole + 32 g/mole= 239 g/mole
So, by stoichiometry they react and are produced:
- H₂S: 1 mole* 34 g/mole= 34 g
- PbS: 1 mole* 239 g/mole= 239 g
Then the following rule of three can be applied: if 239 grams of PbS are produced by stoichiometry from 34 grams of H₂S, 18 grams of PbS from how much mass of H₂S is produced?
mass of H₂S= 2.56 grams
<u><em>2.56 grams of H₂S is needed to produce 18.00g of PbS if the H2S is reacted with an excess (unlimited) supply of Pb(CH₃COO)₂</em></u>
0.563 is a correct value of molar fraction of chloroform.
<u>Explanation:</u>
Mole fraction is a unit of concentration, defined to be equal to the number of moles of a component divided by the total number of moles of a solution. Because it is a ratio, mole fraction is a unit less expression. The mole fraction of all components of a solution, when added together, will equal 1.
Mole fraction is another way of expressing the concentration of a solution or mixture. It is equal to the moles of one component divided by the total moles in the solution or mixture. Mole fraction is used in a variety of calculations, but most notably for calculating partial pressures.
Answer:
Option A. 0.378M
Explanation:
Data obtained from the question include:
Molarity of acid (Ma) =..?
Volume of acid (Va) = 37.0 mL
Volume of base (Vb) = 56.0 mL
Molarity of base (Mb) = 0.250 M
Next, we shall write the balanced equation for the reaction. This is given below:
HCl + NaOH —> NaCl + H2O
From the balanced equation above,
The mole ratio of the acid (nA) = 1
The mole ratio of the base (nB) = 1
Finally, we can determine the molarity of the acid as shown below :
MaVa/MbVb = nA/nB
Ma x 37 / 0.25 x 56 = 1
Cross multiply
Ma x 37 = 0.25 x 56
Divide both side by 37
Ma = 0.25 x 56 /37
Ma = 0.378M
Therefore, the molarity of the acid, HCl is 0.378M
