Answer:
ΔH=15000
J = 15KJ
Explanation:
In this exercise you have find the enthalpy of reaction this is the difference between enthalpy of reactans and products,
For the following equation
H2A(aq) + 2 BOH(aq) → B2A(aq) + 2 H2O(l)
We know that 0.20 moles of BOH reacted with excess amount of H2A solution and 1500. J
so,
(2mol/0,2mol)*1500J=15000J
for de reactions exothermics tha enthalpy is negative so:
ΔH=15000
J = 15KJ
Answer:
A) Dilute the unknown so that it will have an absorbance within the standard curve. Once the diluted unknown concentration is determined, the full strength concentration can be calculated if the dilution process is recorded. Beer's law only applies to dilute solutions, so diluting the unknown is better than making new standards.
Explanation:
Beer's law states that <em>absorbance is proportional to the concentrations of the absorbing species</em>. This is verified in the case of diluted solutions (0≤0.01 M) of most substances. <u>As a solution gets more concentrated, solute molecules interact between themselves because of their proximity. </u>When a molecule interacts with another, the change in their electric properties (including absorbance) is probable. That's why <u>the plot of absorbance versus concentration stops being a straight line</u>, and <u>Beer's law is no longer valid.</u>
Therefore, if the absorbance value is higher than the highest standard, dilutions should be made. Once this concentration is determined, the full strength concentration can be calculated with the inverse of the dilution.
The answer to this question is False
The change in enthalpy associated with the change in the water’s temperature is 1254 J.
<h3>What is specific heat?</h3>
The amount of heat required to increase the temperature of one gram of a substance by one Celsius degree is known as specific heat.
Enthalpy change will be calculated as:
ΔH = -cmΔT, where
m = mass of water = 50g
c = specific heat of water = 4.18J/g°C
ΔT = change in temperature = 28 - 22 = 6 °C
On putting values in the above equation, we get
ΔH = -(4.18)(50)(6) = -1254 J
Hence change in enthalpy of the reaction is -1254 J.
To know more about enthalpy change, visit the below link:
brainly.com/question/11628413
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