Answer:
3.51 g of oxygen per gram of gasoline is required.
Explanation:
Solution:
First of all we will write the balance chemical equation.
C8H18 + 12.5O2 → 8CO2 + 9H2O
This equation shows that,
1 mole of gasoline react with 12.5 mole of oxygen for complete burning.
mass of one mole of gasoline = 8×12 + 18×1 = 114 g
mass of 12.5 mole of oxygen = 12.5 (16×2) = 400 g
Formula:
mass of oxygen per gram of gasoline = (400 / 114) = 3.51
so, 3.51 g of oxygen require for per gram of gasoline.
I believe it's alkali metals and alkaline earth metals.
According to VSEPR theory, the structure of the ammonia molecule, NH3, is linear.
So your answer is B) linear
Answer:
Rate of reaction =
Rate of consumption of A = 
Rate of consumption of B =
Rate of formation of D = 
Explanation:
According to laws of mass action for the given reaction,
![Rate= -\frac{1}{2}\frac{\Delta [A]}{\Delta t}=-\frac{\Delta [B]}{\Delta t}=\frac{1}{2}\frac{\Delta [C]}{\Delta t}=\frac{1}{3}\frac{\Delta [D]}{\Delta t}](https://tex.z-dn.net/?f=Rate%3D%20-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7B%5CDelta%20%5BA%5D%7D%7B%5CDelta%20t%7D%3D-%5Cfrac%7B%5CDelta%20%5BB%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7B%5CDelta%20%5BC%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B1%7D%7B3%7D%5Cfrac%7B%5CDelta%20%5BD%5D%7D%7B%5CDelta%20t%7D)
where, ![-\frac{\Delta [A]}{\Delta t}](https://tex.z-dn.net/?f=-%5Cfrac%7B%5CDelta%20%5BA%5D%7D%7B%5CDelta%20t%7D)
is rate of consumption of A, ![-\frac{\Delta [B]}{\Delta t}](https://tex.z-dn.net/?f=-%5Cfrac%7B%5CDelta%20%5BB%5D%7D%7B%5CDelta%20t%7D)
is rate of consumption of B, ![\frac{\Delta [C]}{\Delta t}](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20%5BC%5D%7D%7B%5CDelta%20t%7D)
is rate of formation of C and ![\frac{\Delta [D]}{\Delta t}](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20%5BD%5D%7D%7B%5CDelta%20t%7D)
is rate of formation of D
Here ![\frac{\Delta [C]}{\Delta t}=2.7mol.dm^{-3}.s^{-1}](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20%5BC%5D%7D%7B%5CDelta%20t%7D%3D2.7mol.dm%5E%7B-3%7D.s%5E%7B-1%7D)
So, Rate of reaction = 
Rate of formation of D = ![(\frac{3}{2}\times \frac{\Delta [C]}{\Delta t})=(\frac{3}{2}\times 2.7mol.dm^{-3}.s^{-1})=4.15mol.dm^{-3}.s^{-1}](https://tex.z-dn.net/?f=%28%5Cfrac%7B3%7D%7B2%7D%5Ctimes%20%5Cfrac%7B%5CDelta%20%5BC%5D%7D%7B%5CDelta%20t%7D%29%3D%28%5Cfrac%7B3%7D%7B2%7D%5Ctimes%202.7mol.dm%5E%7B-3%7D.s%5E%7B-1%7D%29%3D4.15mol.dm%5E%7B-3%7D.s%5E%7B-1%7D)
Rate of consumption of A = ![(\frac{2}{2}\times \frac{\Delta [C]}{\Delta t})=(\frac{2}{2}\times 2.7mol.dm^{-3}.s^{-1})=2.7mol.dm^{-3}.s^{-1}](https://tex.z-dn.net/?f=%28%5Cfrac%7B2%7D%7B2%7D%5Ctimes%20%5Cfrac%7B%5CDelta%20%5BC%5D%7D%7B%5CDelta%20t%7D%29%3D%28%5Cfrac%7B2%7D%7B2%7D%5Ctimes%202.7mol.dm%5E%7B-3%7D.s%5E%7B-1%7D%29%3D2.7mol.dm%5E%7B-3%7D.s%5E%7B-1%7D)
Rate of consumption of B = ![(\frac{1}{2}\times \frac{\Delta [C]}{\Delta t})=(\frac{1}{2}\times 2.7mol.dm^{-3}.s^{-1})=1.35mol.dm^{-3}.s^{-1}](https://tex.z-dn.net/?f=%28%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%5Cfrac%7B%5CDelta%20%5BC%5D%7D%7B%5CDelta%20t%7D%29%3D%28%5Cfrac%7B1%7D%7B2%7D%5Ctimes%202.7mol.dm%5E%7B-3%7D.s%5E%7B-1%7D%29%3D1.35mol.dm%5E%7B-3%7D.s%5E%7B-1%7D)
Answer:
The right answer is "58.3%".
Explanation:
The given values are:
Mass of lithium carbonate,
= 700 mg
i.e.,
= 0.7 g
Mass of tablet,
= 1.20 g
Now,
The percentage of active ingredient will be:
= 
On substituting the given values, we get
= 
= 
= 
%
