Flammmability is a chemical change because it changes the composition of the object. the product is very different from the rectant.
Answer:
2KClO3 -------> 2KCl + 3O2
Explanation:
First, in balancing a chemical reaction such as the one given in the question, you should understand that for an equation to be balanced, the number of atoms and ions on both sides of the equation that is the right and left side must be equal. This follows the law of conservation of mass which tells us that matter can neither be created nor destroyed but can be changed into another form.
Next is to begin balancing the equation by identifying and writing down the substances given:
KCl03 ---------> KCl + O2
Next is to count he number of the individual atoms on each side and find out if they are the same on both sides and if not you must follow the next step.
Add a corresponding number and use it to multiply the atoms involved
KClO3 ---------> KCl + O2
Oxygen is 3 on the left side and two on the other side, so we multiply the left hand side by 2 and the right hand side by 3
2KClO3 -----> KCl + 3O2
The potassium and Chlorine are no longer balanced, so you multiply the right had=nd side of KCl by 2.
2KClO3 -----> 2KCl + 3O2
The reactionis herefore balanced as both sides have equal number of atoms and ions.
Answer : All of the above are valid expressions of the reaction rate.
Explanation :
The given rate of reaction is,
The expression for rate of reaction for the reactant :
![\text{Rate of disappearance of }NH_3=-\frac{1}{4}\times \frac{d[NH_3]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20disappearance%20of%20%7DNH_3%3D-%5Cfrac%7B1%7D%7B4%7D%5Ctimes%20%5Cfrac%7Bd%5BNH_3%5D%7D%7Bdt%7D)
![\text{Rate of disappearance of }O_2=-\frac{1}{7}\times \frac{d[O_2]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20disappearance%20of%20%7DO_2%3D-%5Cfrac%7B1%7D%7B7%7D%5Ctimes%20%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D)
The expression for rate of reaction for the product :
![\text{Rate of formation of }NO_2=+\frac{1}{4}\times \frac{d[NO_2]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20formation%20of%20%7DNO_2%3D%2B%5Cfrac%7B1%7D%7B4%7D%5Ctimes%20%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D)
![\text{Rate of formation of }H_2O=+\frac{1}{6}\times \frac{d[H_2O]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20formation%20of%20%7DH_2O%3D%2B%5Cfrac%7B1%7D%7B6%7D%5Ctimes%20%5Cfrac%7Bd%5BH_2O%5D%7D%7Bdt%7D)
From this we conclude that, all the options are correct.
A sodium ion symbol is written as Na^+.
