The rate law for this reaction is [A]².
Balanced chemical reaction used in this experiment: A + B → P
The reaction rate is the speed at which reactants are converted into products.
Comparing first and second experiment, there is no change in initial rate. The concentration of reactant B is increased by double. Initial rate does not depands on concentration of reactant B.
Comparing first and third experiment, initial rate is nine times greater, while concentration of reactant A is three times greater. Conclusion is that concentration of reactant A is squared and the rate is [A]².
More info about rate law: brainly.com/question/16981791
#SPJ4
Burning Mg in the air and reacting with O2 forming a white powder of MnO
So the equation is going to be:
Mn + O2 ⇒ MnO (this equation is not conserved)
to make it equilibrium:
1- First we should put 2Mno to equal the O2 on both sides.
So it will be:
Mg + O2⇒ 2MgO
2- Second we should put 2Mn to equal the Mn on both sides.
2Mg + O2⇒ 2MgO (this equation is conserved)
After putting the physical states the final equilibrium equation is going to be:
Δ
2Mg(s) + O2(g)⇒ 2MgO(s)
Percent error (%)= 
Accepted value is true value.
Measured values is calculated value.
In the question given Accepted value (true value) = 63.2 cm
Given Measured(calculated values) = 63.1 cm , 63.0 cm , 63.7 cm
1) Percent error (%) for first measurement.
Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.1 cm
Percent error (%)=
Percent error = 0.158 %
2) Percent error (%) for second measurement.
Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.0 cm
Percent error (%)=
Percent error = 0.316 %
3) Percent error (%) for third measurement.
Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.7 cm
Percent error (%)=
Percent error = 0.791 %
Percent error for each measurement is :
63.1 cm = 0.158%
63.0 cm = 0.316%
63.7 cm = 0.791%
Answer:
87.78 in cubed
Explanation:
excuse me if I am wrong i tried my best
don't know if its right
