Answer:
4.56 X 10^ -4 g/mL
Explanation:
A solution is prepared by diluting 6.0 mL of a 7.6 x 10-4 g/mL solution to a total volume of 10.0 mL. Calculate the concentration of the dilute solution.
(7.6 X10^-4 gm/m L) x( 6.0 m L ) = 45.6 X 10^-4 g
this is dissolved )in 10 m L=45.6 X 10^-4 g/ 10
4.56 X 10^ -4 g/mL
check
6/10 =0.6
4.56/7.6 = o.,6
Anything less that 7, would be a base.
Answer:
(a) 7.11x10⁻⁴ M/s
(b) 2.56 mol.L⁻¹.h⁻¹
Explanation:
(a) The reaction is:
O₃(g) + NO(g) → O₂(g) + NO₂(g) (1)
The reaction rate of equation (1) is given by:
(2)
<u>We have:</u>
k: is the rate constant of reaction = 3.91x10⁶ M⁻¹.s⁻¹
[O₃]₀ = 2.35x10⁻⁶ M
[NO]₀ = 7.74x10⁻⁵ M
Hence, to find the inital reacion rate we will use equation (2):
Therefore, the inital reaction rate is 7.11x10⁻⁴ M/s
(b) The number of moles of NO₂(g) produced per hour per liter of air is:
t = 1 h
V = 1 L
![\frac{\Delta[NO_{2}]}{\Delta t} = rate](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%5BNO_%7B2%7D%5D%7D%7B%5CDelta%20t%7D%20%3D%20rate)
![\frac{\Delta[NO_{2}]}{\Delta t} = 7.11 \cdot 10^{-4} M/s*\frac{3600 s}{1 h} = 2.56 mol.L^{-1}.h{-1}](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%5BNO_%7B2%7D%5D%7D%7B%5CDelta%20t%7D%20%3D%207.11%20%5Ccdot%2010%5E%7B-4%7D%20M%2Fs%2A%5Cfrac%7B3600%20s%7D%7B1%20h%7D%20%3D%202.56%20mol.L%5E%7B-1%7D.h%7B-1%7D)
Hence, the number of moles of NO₂(g) produced per hour per liter of air is 2.56 mol.L⁻¹.h⁻¹
I hope it helps you!
Answer:
8 should be placed in front of SO₂.
Explanation:
The easiest way to solve this question is by writing the <u>entire</u> balanced equation:
4FeS₂ + 11O₂ -> 2Fe₂O₃ + 8SO₂
We can achieve this by first balancing the Fe, then S, and finally the O.
We can also double check our answer by counting the number of each element on both sides:
-Reactants: 4 Fe, 8 S, 22 O
-Products: 4 Fe 8 S, 22 O
Since they match, our equation is balanced, and the coefficient in front of SO₂ is 8.