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Studentka2010 [4]
3 years ago
15

How does the law of conservation of mass relate to the number of atoms of each element that are present before a reaction vs. th

e number of atoms of each element that are present after a chemical reaction?
Chemistry
1 answer:
TiliK225 [7]3 years ago
6 0
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.

According to the Law of Conservation, all atoms of the reactant(s) must equal the atoms of the product(s).
As a result, we need to balance chemical equations. We do this by adding in coefficients to the reactants and/or products. The compound(s) itself/themselves DOES NOT CHANGE.
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If 2.0 g of copper(II) chloride react with excess sodium nitrate, what mass of sodium chloride is formed in this double replacem
cluponka [151]

Taking into account the reaction stoichiometry, 1.729 grams of NaCl is formed.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

CuCl₂ + 2 NaNO₃ → Cu(NO₃)₂ + 2 NaCl

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • CuCl₂: 1 mole
  • NaNO₃: 2 moles
  • Cu(NO₃)₂ : 1 mole
  • NaCl: 2 moles

The molar mass of the compounds is:

  • CuCl₂: 134.44 g/mole
  • NaNO₃: 85 g/mole
  • Cu(NO₃)₂ : 187.54 g/mole
  • NaCl: 58.45 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • CuCl₂: 1 mole ×134.44 g/mole= 134.44 grams
  • NaNO₃: 2 moles ×85 g/mole= 170 grams
  • Cu(NO₃)₂ : 1 mole ×187.54 g/mole= 187.54 grams
  • NaCl: 2 moles ×58.45 g/mole= 116.9 grams

<h3>Mass of NaCl formed</h3>

The following rule of three can be applied: If by stoichiometry of the reaction 134.44 grams of CuCl₂ form 116.9 grams of NaCl, 2 grams of CuCl₂ form how much mass of NaCl?

mass of NaCl=\frac{2 grams of CuCl_{2}x116.9 grams of NaCl }{134.44grams of CuCl_{2}}

<u><em>mass of NaCl= 1.739 grams</em></u>

Finally, 1.729 grams of NaCl is formed.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

6 0
1 year ago
2) Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.800 mole of CO and 2.40 mole of H2
nadezda [96]

Answer:

Kc = 3.90

Explanation:

CO reacts with H_2 to form CH_4 and H_2O. balanced reaction is:

CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g)  +  H_2O(g)

No. of moles of CO = 0.800 mol

No. of moles of H_2 = 2.40 mol

Volume = 8.00 L

Concentration = \frac{Moles}{Volume\ in\ L}

Concentration of CO = \frac{0.800}{8.00} = 0.100\ mol/L

Concentration of H_2 = \frac{2.40}{8.00} = 0.300\ mol/L

                 CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g)  +  H_2O(g)

Initial            0.100      0.300             0   0

equi.            0.100 -x    0.300 - 3x     x    x

It is given that,

at equilibrium H_2O (x) = 0.309/8.00 = 0.0386 M

So, at equilibrium CO = 0.100 - 0.0386 = 0.0614 M

At equilibrium H_2 = 0.300 - 0.0386 × 3 = 0.184 M

At equilibrium CH_4 = 0.0386 M

Kc=\frac{[H_2O][CH_4]}{[CO][H_2]^3}

Kc=\frac{0.0386 \times 0.0386}{(0.184)^3 \times 0.0614} =3.90

8 0
3 years ago
A city continuously disposes of effluent from a wastewater treatment plant into a river. The minimum flow in the river is 130 m3
Vera_Pavlovna [14]

Answer:

2.54\ \text{mg/L}

Explanation:

C = Allowable concentration = 1.1 mg/L

Q_1 = Flow rate of river = 130\ \text{m}^/\text{s}

Q_2 = Discharge from plant = 37\ \text{m}^3/\text{s}

C_1 = Background concentration = 0.69 mg/L

C_2 = Maximum concentration that of the pollutant

The concentration of the mixture will be

C=\dfrac{Q_1C_1+Q_2C_2}{Q_1+Q_2}\\\Rightarrow C_2=\dfrac{C(Q_1+Q_2)-Q_1C_1}{Q_2}\\\Rightarrow C_2=\dfrac{1.1(130+37)-130\times 0.69}{37}\\\Rightarrow C_2=2.54\ \text{mg/L}

The maximum concentration that of the pollutant (in mg/L) that can be safely discharged from the wastewater treatment plant is 2.54\ \text{mg/L}.

6 0
2 years ago
Match the prefix with the power of ten it represents.
Sunny_sXe [5.5K]

kilo 103 is the correct answer so it would mostly be 2

6 0
3 years ago
Read 2 more answers
How to find electronegativity differences
lidiya [134]
To calculate electronegativity, start by going online to find an electronegativity table. You can then assess the quality of a bond between 2 atoms by looking up their electronegativities on the table and subtracting the smaller one from the larger one. If the difference is less than 0.5, the bond is nonpolar covalent.
8 0
2 years ago
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