Answer:
Specific heat of substance = 0.897 J/g.°C
Explanation:
Given data:
Specific heat of substance = ?
Mass of substance = 25.0 g
Heat absorbed = 493.4 J
Initial temperature = 12.0 °C
Final temperature = 34°C
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 34°C -12.0°C
ΔT = 22°C
493.4 J = 25 g ×c× 22°C
493.4 J = 550 g.°C ×c
c = 493.4 J / 550 g.°C
c = 0.897 J/g.°C
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
Answer:
The solution to the given problem is done below.
Explanation:
a)
i) =( 0.002 μg / L )( 1mg / 1000 μg )( 1L / kg )( 1000 mil / 1 billion) = 0.002 ppb
ii) =( 0.002 μg / L )( 1mg / 1000 μg )( 1L / kg )( 1,000,000 mil / 1 trillion) = 2 ppt
iii) =( 0.002 μg / L )( 1 mole / 540g ) = 3.7 x 
μM.
b)
i) =( 0.002 μg / g ) = 0.002 ppm
ii) In solids, ppb = μg/kg
=( 0.002 μg / g )( 1000 mil/ 1 billion) = 2 ppb
Addition of water to an alkyne gives a keto‑enol tautomer product and that is the product changed into 2-pentanone, then the alkyne need to had been 1-pentyne. 2-pentyne might have given a combination of 2- and 3-pentanone.
<h3>
What is the keto-enol means in tautomer?</h3>
They carries a carbonyl bond even as enol implies the presence of a double bond and a hydroxyl group. The keto-enol tautomerization equilibrium is depending on stabilization elements of each the keto tautomer and the enol tautomer.
- The enol that could provide 2-pentanone might had been pent-1- en - 2 -ol. Because an equilibrium favors the ketone so greatly, equilibrium isn't an excellent description.
- If the ketone have been handled with bromine, little response might be visible because the enol content material might be too low.
- If a catalyst have been delivered, NaOH for example, then formation of the enolate of pent-1-en - 2 - ol might shape and react with bromine.
- This might finally provide a bromoform product. Under acidic conditions, the enol might desire formation of the greater substituted enol constant with alkene stability.
