If the O-Zone Tropo increased to a point, like it has, it can cause a huge variety of health problems and even death! Some of the problems include, but are not limited to: Asthma, E<span>mphysema, Chest Pain, Inflamed Lungs and Lung Scarring from tissue damage. </span>
Answer:
4
Explanation:
Speed=Distance/time, so 20/5 divide time divided by distance. so the answer is 4.
Explanation:
It is known that equation for ideal gas is as follows.
PV = nRT
The given data is as follows.
Pressure, P = 1500 psia, Temperature, T = 
= 104 + 460 = 564 R
Volume, V = 2.4 cubic ft, R = 10.73 
Also, we know that number of moles is equal to mass divided by molar mass of the gas.
n = 
m = 
= 
= 9.54 lb
Hence, molecular weight of the gas is 9.54 lb.
- We will calculate the density as follows.
d = 
= 
= 3.975 
- Now, calculate the specific gravity of the gas as follows.
Specific gravity relative to air = 
= 
= 51.96
Answer:
Explanation:
E = (hc)/(λ)
E = (6.624x10^(-27))Js x ((3×10^8)ms^(-1)) /
(77.8x10^(-9)m)
E = 2.55 x 10^(-11) J
Answer:
Percent yield = 89.1%
Explanation:
Based on the equation:
Cl₂ + 2KI → 2KCl + I₂
<em>1 mole of Cl₂ reacts with 2 moles of KI to produce to moles of KCl</em>
<em />
To solve this quesiton we must find the moles of each reactant in order to find the limiting reactant. With the limiting reactant we can find the moles of KCl and the mass:
<em>Moles Cl₂:</em>
8x10²⁵ molecules * (1mol / 6.022x10²³ molecules) = 133 moles
<em>Moles KI -Molar mass: 166.0028g/mol-</em>
25g * (1mol / 166.0028g) = 0.15 moles
Here, clarely, the KI is the limiting reactant
As 2 moles of KI produce 2 moles of KCl, the moles of KCl produced are 0.15 moles. The theoretical mass is:
0.15 moles * (74.5513g / mol) =
11.2g KCl
Percent yield is: Actual yield (10.0g) / Theoretical yield (11.2g) * 100
<h3>Percent yield = 89.1%</h3>
