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Nookie1986 [14]
3 years ago
6

why is the law of conservation of mass important to understand when writing chemical formulas and chemical equations​

Chemistry
1 answer:
Naddika [18.5K]3 years ago
4 0

Explanation:

The law of conservation of mass is important to understand when writing chemical equations because it is very useful in making chemical calculations on reactants and the products.

  The law states that "the mass of products in chemical reactions is equal to the mass of the reactants".

  • The law shows that new atoms are not created in a reaction, bonds are just broken and rearranged and new compounds are formed.
  • Based on this premise, it is easy to track the amount of products that formed.
  • The law is also an indicator of how efficient a reaction can be.
  • Before a reaction starts, the law can be used to predict the amount and nature of products that will form.

Learn more:

Chemical laws brainly.com/question/5896850

#learnwithBrainly

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What is the mass of 0.0490 mole of potassium sulfate?
Angelina_Jolie [31]

Answer:

8.54g

Explanation:

we know that: n=m/Mm

where:

  • n represents moles
  • m represent mass
  • and Mm represents molar mass

rearrange for m: m= n x Mm

m= 0.0490mol x 174.259g/mol

m= 8.54g

7 0
3 years ago
A chemist designs a galvanic cell that uses these two half-reactions:
Nonamiya [84]

Answer :

(a) Reaction at anode (oxidation) : 4Fe^{2+}\rightarrow 4Fe^{3+}+4e^-  

(b) Reaction at cathode (reduction) : O_2+4H^++4e^-\rightarrow 2H_2O  

(c) O_2+4H^++4Fe^{2+}\rightarrow 2H_2O+4Fe^{3+}

(d) Yes, we have have enough information to calculate the cell voltage under standard conditions.

Explanation :

The half reaction will be:

Reaction at anode (oxidation) : Fe^{2+}\rightarrow Fe^{3+}+e^-     E^0_{anode}=+0.771V

Reaction at cathode (reduction) : O_2+4H^++4e^-\rightarrow 2H_2O     E^0_{cathode}=+1.23V

To balance the electrons we are multiplying oxidation reaction by 4 and then adding both the reaction, we get:

Part (a):

Reaction at anode (oxidation) : 4Fe^{2+}\rightarrow 4Fe^{3+}+4e^-     E^0_{anode}=+0.771V

Part (b):

Reaction at cathode (reduction) : O_2+4H^++4e^-\rightarrow 2H_2O     E^0_{cathode}=+1.23V

Part (c):

The balanced cell reaction will be,

O_2+4H^++4Fe^{2+}\rightarrow 2H_2O+4Fe^{3+}

Part (d):

Now we have to calculate the standard electrode potential of the cell.

E^o=E^o_{cathode}-E^o_{anode}

E^o=(1.23V)-(0.771V)=+0.459V

For a reaction to be spontaneous, the standard electrode potential must be positive.

So, we have have enough information to calculate the cell voltage under standard conditions.

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3 years ago
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Answer:

It is double displacement reaction

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