To convert the formula unit to mass, we need to divide the given formula units by Avogadro's number, 6.022 x 10^23 and we get the mole of beryllium nitrate. To convert to mass, we need to multiply the number of moles with the molecular formula of the compound which is 133.022 g/mol.The answer is 0.006185 g or 6.185 mg.
Answer:
This effect is caused by increased levels of carbon dioxide, chlorofluorocarbons and other gases in the air, many of them released by human activity. greenhouse effect The warming of Earth's atmosphere due to the buildup of heat-trapping gases, such as carbon dioxide and methane.
Explanation:
The average speed :
1. 10.44 m/s
2. 10.42 m/s
3. 9.26 m/s
The distance 100 m have the greatest average speed
<h3>Further explanation
</h3>
Given
Distance and time of runner
Required
Average speed
Solution
<em>
Average speed : total distance : total time
</em>
1. d = 100 m, t = 9.58 s
Average speed : 100 : 9.58 = 10.44 m/s
2. d=200 m, t=19.19 s
Average speed : 200 : 19.19 = 10.42 m/s
3. d=400 m, t = 43.18 s
Average speed : 400 : 43.18 = 9.26 m/s
The distance 100 m have the greatest average speed
Explanation:
1) Thermal and Electric conductivity
2) Metallic strength
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:
![rate = k*[O_{3}][NO]](https://tex.z-dn.net/?f=%20rate%20%3D%20k%2A%5BO_%7B3%7D%5D%5BNO%5D%20)
(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):
![rate = k*[O_{3}]_{0}[NO]_{0} = 3.91 \cdot 10^{6} M^{-1}s^{-1}*2.35\cdot 10^{-6} M*7.74 \cdot 10^{-5} M = 7.11 \cdot 10^{-4} M/s](https://tex.z-dn.net/?f=%20rate%20%3D%20k%2A%5BO_%7B3%7D%5D_%7B0%7D%5BNO%5D_%7B0%7D%20%3D%203.91%20%5Ccdot%2010%5E%7B6%7D%20M%5E%7B-1%7Ds%5E%7B-1%7D%2A2.35%5Ccdot%2010%5E%7B-6%7D%20M%2A7.74%20%5Ccdot%2010%5E%7B-5%7D%20M%20%3D%207.11%20%5Ccdot%2010%5E%7B-4%7D%20M%2Fs%20)
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!
