It will take 1.11 min to heat the sample to its melting point.
Melting point = - 20°C
Boiling point = 85°C
∆H of fusion = 180 J/g
∆H of vap = 500 J/g
C(solid) = 1.0 J/g °C
C(liquid) = 2.5 J/g °C
C(gas) = 0.5 J/g °C
Mass of sample = 25 g
Initial temperature = - 40°C
Final temperature = 100°C
Rate of heating = 450 J/min
Specific heat capacity formula:- q = m ×C×∆T
Here, q = heat energy
m = mass
C = specific heat
∆T = temperature change
Melting point = - 20°C
C(solid) = 1.0 J/g °C
∆T = final temperature - initial temperature = -20 - (-40) = 20
Put these value in Specific heat capacity formula
q = m ×C×∆T
q = 25×1.0×20
=500J
The Rate of heating = 450 J/min
i.e. 450J = 1min
so, 500J = 1.11min
1.11 minutes does it take to heat the sample to its melting point.
The specific heat capacity is defined as the amount of heat absorbed in line with unit mass of the material whilst its temperature increases 1 °C.
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Species that have a lone pair of electrons often donate electrons by resonance while substituents that are electron deficient take away electrons by resonance.
<h3>What is resonance?</h3>
The term resonace has to do with the movement of electron pairs in a molecule. Inductive effects has to do with the drawing of electron density towards an atom or bond.
The two effects depends on the nature of a substituent. For instance, species that have a lone pair of electrons often donate electrons by resonance while substituents that are electron deficient take away electrons by resonance.
The question is incomplete hence the exact nature of the substituents can not be determined.
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Answer:
Q=mcΔT
Explanation:
The formula for expressing the amount of heat transferred between energy stores is given by the equation. The specific heat capacity of water is 4180 J/kgoC (Joules per kilogram per degree), this means it takes 4180 J of heat energy to raise the temperature of 1 kg of water by 1oC.
The term sensitivity in Analytical Chemistry is "the slope of the calibration curve or a function of analyte concentration or amount".
<u>Answer:</u> Option B
<u>Explanation:</u>
In a sample, the little amounts of substances can be accurately evaluated by a method is termed as "Analytical sensitivity". This detect a target analyte like an antibody or antigen, process is considered as potential of a test to and generally demonstrated as the analyte's minimum detectable concentration.
The acceptable diagnostic sensitivity is not guaranteed by high analytical sensitivity. The percentage of individuals who have a given disarray who are identified by the method as positive for the disarray is known as "Diagnostic sensitivity".
<h3>
Answer:</h3>
82.11%
<h3>
Explanation:</h3>
We are given;
- Theoretical mass of the product is 137.5 g
- Actual mass of the product is 112.9 g
We are supposed to calculate the percentage yield
- We need to know how percentage yield is calculated;
- To calculate the percentage yield we get the ratio of the actual mass to theoretical mass and express it as a percentage.
Thus;
% yield = (Actual mass ÷ Experimental mass) × 100%
= (112.9 g ÷ 137.5 g) × 100%
= 82.11%
Therefore, the percentage yield of the product is 82.11 %
