To balance a reaction, we must say to it that the number of elements in one side is equal to the other side. For a combustion reaction such as the one given, we first need to balance the number of carbon atoms, then the hydrogen atoms and lastly the oxygen atoms.
<span>C4H10 + 13/2O2 → 4CO2 + 5H2O.</span>
Answer:
It is valid because his experiment had clear variables.
Explanation:
He had clear variables and did everyhting the same for both plants, even though he had calcium in one and not in the other, which gives him a true and good result.
Answer:
44.6millilitres
Explanation:
Using the general gas law equation as follows:
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (atm)
V1 = initial volume (L)
T1 = initial temperature (K)
P2 = final pressure (atm)
V2 = final volume (L)
T2 = final temperature (K)
According to this question;
V1 = 30mL
T1 = 273K (STP)
P1 = 1 atm (STP)
V2 = ?
T2 = 300K
P2 = 75.0 kPa = 75 × 0.00987 = 0.74atm
Using P1V1/T1 = P2V2/T2
1 × 30/273 = 0.74×V2/300
30/273 = 0.74V2/300
Cross multiply
300 × 30 = 273 × 0.74V2
9000 = 202.02V2
V2 = 9000/202.02
V2 = 44.55
V2 = 44.6millilitres.
Answer : The concentration of 
and 
are 
and 
respectively.
Solution : Given,
pH = 4.10
pH : pH is defined as the negative logarithm of hydronium ion concentration.
Formula used : ![pH=-log[H_3O^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH_3O%5E%2B%5D)
First we have to calculate the hydronium ion concentration by using pH formula.
![4.10=-log[H_3O^+]](https://tex.z-dn.net/?f=4.10%3D-log%5BH_3O%5E%2B%5D)
![[H_3O^+]=antilog(-4.10)](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3Dantilog%28-4.10%29)
![[H_3O^+]=7.94\times 10^{-5}](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D7.94%5Ctimes%2010%5E%7B-5%7D)
Now we have to calculate the pOH.
As we know, 
Now we have to calculate the hydroxide ion concentration.
![pOH=-log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-log%5BOH%5E-%5D)
![9.9=-log[OH^-]](https://tex.z-dn.net/?f=9.9%3D-log%5BOH%5E-%5D)
![[OH^-]=antilog(-9.9)](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3Dantilog%28-9.9%29)
![[OH^-]=1.258\times 10^{-10}](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.258%5Ctimes%2010%5E%7B-10%7D)
Therefore, the concentration of and are and respectively.
