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lesantik [10]
3 years ago
7

All chemical reactions have a conservation of(1) mass, only(2) mass and charge, only(3) charge and energy, only(4) mass, charge,

and energy
Chemistry
2 answers:
Semenov [28]3 years ago
7 0
All chemical reactions have a conversation of mass and energy. 

Because:- There are only two laws for conversation in a chemical reaction. The conversation of mass, no mass can be created nor destroyed. Also, the law of conversation of mass states that no energy can be created nor destroyed in a chemical reaction. The charge can obviously change because they can bond and change charges. 
tankabanditka [31]3 years ago
3 0

<u>Answer:</u> The correct answer is option 4.

<u>Explanation: </u>

<u>Law of conservation of mass</u> states that mass can neither be created nor be destroyed, but it can only be transformed from one form to another. It always remain conserved. In a chemical reaction, mass on the reactant side will be equal to the mass on the product side.

For Example: Reaction of hydrogen and oxygen to form water.

2H_2+O_2\rightarrow 2H_2O

Mass on the reactant side: [tex][2(2\times 1)+(2\times 16)]=36g/mol

Mass on the product side: [2\times ((2\times 1)+16)]=36g/mol

<u>Law of conservation of energy</u> states that energy can neither be created not be destroyed, but it can only be transformed from one form to another. In a chemical reaction, energy given to the reactants is released in the forma of energy of the products.

For Example: Melting of ice. In this process, the energy given to the ice is released in the form of kinetic energy of the water molecules (product).

<u>Law of Conservation of charge</u> states that charge can neither be created not be destroyed, but it can only be transformed from one form to another. In a chemical reaction, total charge on the reactant side will always be equal to the total charge on the product side.

For Example: Reduction of chlorine gas. The chemical equation follows:

Cl_2+2e^-\rightarrow 2Cl^-

Total charge on the reactant side = -2

Total charge on the product side = -2

From the above information, the correct answer is Option 4.

You might be interested in
antimony has two naturally occuring isotopes, sb121sb121 and sb123sb123 . sb121sb121 has an atomic mass of 120.9038 u120.9038 u
Luda [366]

Considering the definition of atomic mass, isotopes and atomic mass of an element, sb121 has a percent natural abundance of 0.5726 or 57.26% and sb123 has a percent natural abundance of 0.4284 or 42.84%.

<h3>Definition of atomic mass</h3>

The atomic mass is obtained by adding the number of protons and neutrons in a given nucleus of a chemical element.

<h3>Definition of isotope</h3>

Isotopes are the chemical elements in which atomic numbers are the same, but the number of neutrons is different.

<h3>Definition of atomic mass</h3>

The atomic mass of an element is the weighted average mass of its natural isotopes.

This is, the atomic masses of elements are usually calculated as the weighted average of the masses of the different isotopes of each element, considering the relative abundance of each of them.

<h3>Percent natural abundance of each isotope</h3>

In this case, antimony has two naturally occuring isotopes, sb121 and sb123. You know:

  • sb121 has an atomic mass of 120.9038 u.
  • sb121 has a percent natural abundance of x.
  • sb123 has an atomic mass of 122.9042 u.
  • sb123 has a percent natural abundance of 1 -x (or, what is the same, the abundance is 100% - x%, since both isotopes form 100% of the element.)
  • Antimony has an average atomic mass of 121.7601 u

The average mass of antimony is expressed as:

121.7601 u= 120.9038 u x + 122.9042 u× (1 -x)

Solving:

121.7601 u= 120.9038 u x + 122.9042 u - 122.9042 u x

121.7601 u - 122.9042 u= 120.9038 u x - 122.9042 u x

(-1.1441 u)= (-2.0014) x

(-1.1441 u)÷ (-2.0014)= x

<u><em>0.5726= x or 57.26%</em></u>

So, 1 -x= 1- 0.5716 → <u><em>1-x= 0.4284 or 42.84%</em></u>

<u><em /></u>

Finally, sb121 has a percent natural abundance of 0.5726 or 57.26% and sb123 has a percent natural abundance of 0.4284 or 42.84%.

Learn more about average atomic mass:

brainly.com/question/4923781

brainly.com/question/1826476

brainly.com/question/15230683

brainly.com/question/7955048

#SPJ1

5 0
2 years ago
Consider the reaction below. NH4+ + H2O mc013-1.jpg NH3 + H3O+ Which is an acid-conjugate base pair? NH4+ and NH3 NH4+ and H3O+
juin [17]
NH4+ and NH3 are an acid-conjugate base pair, since NH4+ is an acid, while NH3 is its conjugate base (since it is without the H+).
H2O and H3O+ can also be considered an acid-conjugate base pair, since H3O+ is an acid, while H2O would be its conjugate base. (But if only 1 answer is to be selected, it should be the NH4+ and NH3)
NH4+ and H3O+ are both acids, and both H2O and NH3 can be considered bases.
4 0
3 years ago
Read 2 more answers
Can 1750 mL of water dissolve 4.6 moles of Copper Sulfate CuSO4? _________ Why? / Why not?
Wewaii [24]

Answer:

  • <u>No, you cannot dissolve 4.6 moles of copper sulfate, CuSO₄, in 1750mL of water.</u>

Explanation:

This question is part of a Post-Lab exercise sheet.

Such sheet include the saturation concentrations for several salts.

The saturation concentration of Copper Sulfate, CuSO₄, indicated in the table is 1.380M.

That means that 1.380 moles of copper sulfate is the maximum amount that can be dissolved in one liter of solution.

Find the molar concentration for 4.6 moles of copper sulfate in 1,750 mL of water.

You need to assume that the volume of water (1750mL) is the volume of the solution. This is, that the 4.6 moles of copper sulfate have a negligible volume.

<u>1. Volume in liters:</u>

  • V = 1,750 mL × 1 liter / 1,000 mL = 1.75 liter

<u />

<u>2. Molar concentration, molarity, M:</u>

  • M = number of moles of solute / volume of solution in liters

  • M = 4.6 moles / 1.75 liter = 2.6 M

Since the solution is saturated at 1.380M, you cannot reach the 2.6M concentration, meaning that you cannot dissolve 4.6 moles of copper sulfate, CuSO₄ in 1750mL of water.

8 0
3 years ago
In this double replacement reaction, which are the products? hcl + naoh —&gt; ______?
Anarel [89]
The answer is:  " NaCl + H₂O " ; (or; write as:  " H₂O + NaCl " ) .
________________________________________________________
Specifically:
_________________________________________________________

    HCl + NaOH  —>  NaCl + H₂O  ;   or; write as:

    NaOH + HCl   —>  H₂O + NaCl  .
_______________________________________________________
            This type of "double-replacement" reaction is called "neutralization".
 
             Since we are adding a strong acid to a strong base (reactants), we know that the product will be:  1) a salt ; and 2) water.  Since we know one of the reactants will be "water" (H₂O) ; we can find the base (i.e. , the "remaining product") from selecting the "unused elements" to form the corresponding "salt".
________________________________________________________
3 0
3 years ago
Read 2 more answers
what is the percent yield of titanium (II) oxide if 20.0 grams of titanium (II) sulfide is reacted with water? The actual yield
earnstyle [38]

Answer : The percent yield of titanium (II) oxide is, 142.5 % and the impurities could have caused the percent yield to be so high.

Explanation : Given,

Mass of titanium(II) sulfide = 20.0 g

Molar mass of titanium(II) sulfide = 79.9 g/mole

Molar mass of titanium(II) oxide = 63.9 g/mole

First we have to calculate the moles of titanium(II) sulfide.

\text{ Moles of titanium(II) sulfide}=\frac{\text{ Mass of titanium(II) sulfide}}{\text{ Molar mass of titanium(II) sulfide}}=\frac{20.0g}{79.9g/mole}=0.2503moles

Now we have to calculate the moles of titanium(II) oxide.

The balanced chemical reaction is,

TiS+H_2O\rightarrow TiO+H_2S

From the reaction, we conclude that

As, 1 mole of titanium(II) sulfide react to give 1 mole of titanium(II) oxide

So, 0.2503 mole of titanium(II) sulfide react to give 0.2503 mole of titanium(II) oxide

Now we have to calculate the mass of titanium(II) oxide.

\text{ Mass of titanium(II) oxide}=\text{ Moles of titanium(II) oxide}\times \text{ Molar mass of titanium(II) oxide}

\text{ Mass of titanium(II) oxide}=(0.2503moles)\times (63.9g/mole)=15.99g

To calculate the percentage yield of titanium (II) oxide, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of titanium (II) oxide = 22.8 g

Theoretical yield of titanium (II) oxide = 15.99 g

Putting values in above equation, we get:

\%\text{ yield of titanium (II) oxide}=\frac{22.8g}{15.99g}\times 100\\\\\% \text{yield of titanium (II) oxide}=142.5\%

Hence, the percent yield of titanium (II) oxide is, 142.5 %

If the percent yields is greater than 100% that means the product of the reaction contains impurities which cause its mass to be greater than it actually.

5 0
3 years ago
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