Answer:
1.552 moles
Explanation:
First, we'll begin by writing a balanced equation for the reaction showing how C8H18 is burn in air to produce CO2.
This is illustrated below:
2C8H18 + 25O2 -> 16CO2 + 18H2O
Next, let us calculate the number of mole of C8H18 present in 22.1g of C8H18. This is illustrated below:
Molar Mass of C8H18 = (12x8) + (18x1) = 96 + 18 = 114g/mol
Mass of C8H18 = 22.1g
Mole of C8H18 =..?
Number of mole = Mass /Molar Mass
Mole of C8H18 = 22.1/144
Mole of C8H18 = 0.194 mole
From the balanced equation above,
2 moles of C8H18 produced 16 moles of CO2.
Therefore, 0.194 mole of C8H18 will produce = (0.194x16)/2 = 1.552 moles of CO2.
Therefore, 1.552 moles of CO2 are emitted into the atmosphere when 22.1 g C8H18 is burned
Answer:
The pH and pOH of a 2.2*10⁻³ HBr solution is 2.66 and 11.34 respectively.
Explanation:
pH - short for hydrogen potential - is a measure of the acidity or alkalinity of a solution. So the pH is a parameter that indicates the concentration of hydrogen ions [H]⁺ that exist in a solution.
The pH is expressed as the negative base 10 logarithm of the hydrogen ion concentration. This is represented by:
pH= - log [H⁺]
pH is measured on a scale of 0 to 14. On this scale, a pH value of 7 is neutral, which means that the substance or solution is neither acidic nor alkaline. A pH value of less than 7 means that it is more acidic, and a pH value of more than 7 means that it is more alkaline.
HBr is a strong acid. Then, in aqueous solution it will be totally dissociated. So the proton concentration is equal to the initial concentration of acid:
[H⁺]= [HBr]= 2.2*10⁻³ M
So:
pH= - log (2.2*10⁻³)
pH= 2.66
On the other hand, pOH is a measure of the concentration of hydroxyl ions in a solution. The sum of pH and pOH equals 14:
pH + pOH= 14
2.66 + pOH= 14
pOH= 14 - 2.66
pOH= 11.34
<u><em>The pH and pOH of a 2.2*10⁻³ HBr solution is 2.66 and 11.34 respectively.</em></u>
<span>Selenium (se)
........
.....</span>
Explanation:
<h3>Hinsberg reagent is an alternative name for benzene sulfonyl chloride. This name is given for its use in the Hinsberg test for the detection and distinction of primary, secondary, and tertiary amines in a given sample. This reagent is an organosulfur compound.</h3>
Answer:
0.2193 μm
Explanation:
The reaction showing the Photodissociation of ozone (O3) is given below as:
O₃ + hv --------------------------> O₂ + O⁺
H° (142.9) (0) (438kJ/mol).
The first thing to do here is to determine the change in the enthalpy of the total reaction, this can be done by subtracting the change in the enthalpy of the reactant from the change in enthalpy in the product. Hence, we have:
ΔH° = [438 kJ/mol + 247.5 kJ/mol] - (142.9) = 542.6 KJ/mol.
This value, that is 542.6 KJ/mol will then be used in the determination of the value for the maximum wavelength that could cause this photodissociation.
Therefore, the maximum wavelength could cause this photodissociation ≤ h × c/ E = [ 1.199 × 10⁻⁴]/ 542.6 = 2.193 × 10⁻⁷ = 0.2193 μm
