Answer:
The fire may grow bigger, depending on the gas.
Explanation:
If you expose more air to a small flame then it could grow larger because air keeps fire alive.
Answer:
The average rate of the reaction in terms of disappearance of A is 0.0004 M/s.
Explanation:
Average rate of the reaction is defined as ratio of change in concentration of reactant with respect to given interval of time.
![R_{avg}=-\frac{[A]_2-[A]_1}{t_2-t_1}](https://tex.z-dn.net/?f=R_%7Bavg%7D%3D-%5Cfrac%7B%5BA%5D_2-%5BA%5D_1%7D%7Bt_2-t_1%7D)
Where :
= initial concentration of reactant at 
.
= Final concentration of reactant at 
.
2A+3B → 3C+2D
![R_{avg}=-\frac{1}{2}\frac{[A]_2-[A]_1}{t_2-t_1}](https://tex.z-dn.net/?f=R_%7Bavg%7D%3D-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7B%5BA%5D_2-%5BA%5D_1%7D%7Bt_2-t_1%7D)
The concentration of A at (
) = 
The concentration of A at (
) = 
The average rate of reaction in terms of the disappearance of reactant A in an interval of 0 seconds to 20 seconds is :
The average rate of the reaction in terms of disappearance of A is 0.0004 M/s.
Answer:Reflection is the bounce back of the rays of light
Answer:
Explanation:
Let the number of moles of oxygen = x
2H2 + O2 --> 2 H2O
x 13.3
Since the balance number for oxygen is 1 and the balance number for water is 2, you must set up a proportion. (Those balance numbers represent the number of moles).
1/x = 2 / 13.3 Cross Multiply
2*x = 13.3 Divide both sides by 2
2x/2 = 13.3/2
x = 6.65
You need 6.65 moles of oxygen.
Answer:
0.07172 L = 7.172 mL.
Explanation:
- We can use the general law of ideal gas: <em>PV = nRT.
</em>
where, P is the pressure of the gas in atm (P = 1.0 atm, Standard P).
V is the volume of the gas in L (V = ??? L).
n is the no. of moles of the gas in mol (n = 3.2 x 10⁻³ mol).
R is the general gas constant (R = 0.0821 L.atm/mol.K),
T is the temperature of the gas in K (T = 273 K, Standard T).
<em>∴ V = nRT/P =</em> (3.2 x 10⁻³ mol)(0.0821 L.atm/mol.K)(273 K)/(1.0 atm) = <em>0.07172 L = 7.172 mL.</em>
