Answer: 27.09 ppm and 0.003 %.
First, <u>for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. </u>Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.
Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.
So, according to the <em>law of ideal gases,</em>
PV = nRT
where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)
The moles of CO will be,
n = 35 mg x 
x 
→ n = 0.00125 mol
We clear V from the equation and substitute P = 0.92 atm and
T = -30 ° C + 273.15 K = 243.15 K
V = 
→ V = 0.0271 L
As 1000 cm³ = 1 L then,
V = 0.0271 L x 
= 27.09 cm³
<u>Then the acceptable concentration </u><u>c</u><u> of CO in ppm is,</u>
c = 27 cm³ / m³ = 27 ppm
<u>To express this concentration in percent by volume </u>we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:
c = 27.09 
x 
x 100%
c = 0.003 %
So, <u>the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is </u>27.09 ppm and 0.003 %.
<span>Answer: 17.8 cm
</span>
<span>Explanation:
</span>
<span>1) Since temperature is constant, you use Boyle's law:
</span>
<span>PV = constant => P₁V₁ = P₂V₂
</span><span>=> V₁/V₂ = P₂/P₁</span>
<span>
2) Since the ballon is spherical:
</span><span>V = (4/3)π(r)³</span>
<span>
Therefore, V₁/V₂ = (r₁)³ / (r₂)³
</span>
<span>3) Replacing in the equation V₁/V₂ = P₂/P₁:
</span><span><span>(r₁)³ / (r₂)³ </span>= P₂/P₁</span>
<span>
And you can solve for r₂: (r₂)³ = (P₁/P₂) x (r₁)³
</span>(r₂)³ = (1.0 atm / 0.87 atm) x (17 cm)³ = 5,647.13 cm³
<span>
r₂ = 17.8 cm</span>
<u>Answer:</u> The limiting reagent in the reaction is bromine.
<u>Explanation:</u>
Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.
Excess reagent is defined as the reagent which is left behind after the completion of the reaction.
Given values:
Moles of iron = 10.0 moles
Moles of bromine = 12.0 moles
The chemical equation for the reaction of iron and bromine follows:
By the stoichiometry of the reaction:
If 3 moles of bromine reacts with 2 moles of iron
So, 12.0 moles of bromine will react with = 
of iron
As the given amount of iron is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.
Hence, bromine is considered a limiting reagent because it limits the formation of the product.
Thus, the limiting reagent in the reaction is bromine.
The quantity of substance remains after 850 years is 8.98g if the half life of radioactive radium is 1,599 years.
<h3>What is half life period? </h3>
The time taken by substance to reduce to its half of its initial concentration is called half life period.
We will use the half- life equation N(t)
N e^{(-0.693t) /t½}
Where,
N is the initial sample
t½ is the half life time period of the substance
t2 is the time in years.
N(t) is the reminder quantity after t years .
Given
N = 13g
t = 350 years
t½ = 1599 years
By substituting all the value, we get
N(t) = 13e^(0.693 × 50) / (1599)
= 13e^(- 0.368386)
= 13 × 0.691
= 8.98
Thus, we calculated that the quantity of substance remains after 850 years is 8.98g if the half life of radioactive radium is 1,599 years.
learn more about half life period:
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