Answer:
The correct answer to the following question will be "4.08 × 10⁻¹⁹ Joule".
Explanation:
Given:
Wavelength, λ = 486.0 nm
As we know,
On putting the estimated values, we get
⇒ 
⇒ 
∴ 1 ev = 1.6 × 10⁻¹⁹ J
Now,
Energy, 
⇒ 
Generally speaking, organic molecules tend to dissolve in solvents that have similar physical properties. A good rule of thumb is that "like dissolves like". Meaning, polar compounds can dissolve polar compounds and nonpolar compounds can dissolve nonpolar compounds.
To apply this to the current problem, we are told that the brushes are being cleaned with vegetable oil or mineral oil. In this case, the oils are used as solvents. In order for these solvents to be effective, the compounds they are trying to dissolve must be similar in structure and properties to other oils. Therefore, vegetable oil or mineral oil will be most effective in removing oil-based paints, as these will have the similar properties needed to dissolve in the oil solvents.
The solubility of carbon dioxide at 400 kPa at room temperature is ;
( B ) 0.61 CO2/L
<u>Given data </u>
pressure of CO₂ = 400 Kpa = 3.95 atm
Kh of CO₂ = 3.3 * 10⁻² mol/L.atm
<h3>Calculate the solubility of carbon dioxide </h3>
Solubility = pressure * Kh value of CO₂
= 3.95 atm * 3.3 * 10⁻² mol / L.atm
= 0.13 mol/l CO₂
= 0.61 CO₂ / L
Hence we can conclude that the solubility of CO₂ at 400 kPa is 0.13 mol/l CO₂.
Learn more about solubility : brainly.com/question/23946616
From the options the closest answer is ( B ) 0.61 CO₂ / L
Answer:
bacic
Explanation:
it is very soluble in water
Answer:
T2 = 135.1°C
Explanation:
Given data:
Mass of water = 96 g
Initial temperature = 113°C
Final temperature = ?
Amount of energy transfer = 1.9 Kj (1.9×1000 = 1900 j)
Specific heat capacity of aluminium = 0.897 j/g.°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 = T2 - T1
Now we will put the values in formula.
Q = m.c. ΔT
1900 j = 96 g × 0.897 j/g.°C × T2 - 113°C
1900 j = 86.112 j/°C × T2 - 113°C
1900 j / 86.112 j/°C = T2 - 113°C
22.1°C + 113°C = T2
T2 = 135.1°C
