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.
Learn more about specific heat capacity here:- brainly.com/question/26866234
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Among the choices provided, the statement that correctly describes the rusting process below is that "Oxygen was reduced over the course of this reaction" as <span> iron can't be the oxidizing agent, because it is the one being oxidized.
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1 wavelength, 2 crest, 3 trough, 4 wave height <3
The given question is incomplete. The complete question is :
In each row check off the boxes that apply to the underlined reactant. The underlined reactant acts as a... (check all that apply)
1. 
here underlined is 
A. Brønsted-Lowry acid
B. Brønsted-Lowry base
C. Lewis acid
D. Lewis base
2. 
Here underlined is 
A. Brønsted-Lowry acid
B. Brønsted-Lowry base
C. Lewis acid
D. Lewis base
3. 
Here underlined is 
A. Brønsted-Lowry acid
B. Brønsted-Lowry base
C. Lewis acid
D. Lewis base
Answer: 1. Brønsted-Lowry acid
2. Lewis base
3. Brønsted-Lowry base
Explanation:
According to the Bronsted Lowry conjugate acid-base theory, an acid is defined as a substance which donates protons and a base is defined as a substance which accepts protons.
According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs.
1. 
As 
is donating a proton , it acts as a bronsted acid.
2.
As contains a lone pair of electron on nitrogen , it can easily donate electrons to 
and act as lewi base.
3.
As 
is accepting a proton , it acts as a bronsted base.
Answer : The value of 
for the reaction is -959.1 kJ
Explanation :
The given balanced chemical reaction is,
First we have to calculate the enthalpy of reaction 
.
![\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5Bn_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28H_2O%29%7D%2Bn_%7BSO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28SO_2%29%7D%5D-%5Bn_%7BH_2S%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28H_2S%29%7D%2Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28O_2%29%7D%5D)
where,
= enthalpy of reaction = ?
n = number of moles
= standard enthalpy of formation
Now put all the given values in this expression, we get:
![\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5B2mole%5Ctimes%20%28-242kJ%2Fmol%29%2B2mole%5Ctimes%20%28-296.8kJ%2Fmol%29%7D%5D-%5B2mole%5Ctimes%20%28-21kJ%2Fmol%29%2B3mole%5Ctimes%20%280kJ%2Fmol%29%5D)
conversion used : (1 kJ = 1000 J)
Now we have to calculate the entropy of reaction 
.
![\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5Bn_%7BH_2O%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28H_2O%29%7D%2Bn_%7BSO_2%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28SO_2%29%7D%5D-%5Bn_%7BH_2S%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28H_2S%29%7D%2Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28O_2%29%7D%5D)
where,
= entropy of reaction = ?
n = number of moles
= standard entropy of formation
Now put all the given values in this expression, we get:
![\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5B2mole%5Ctimes%20%28189J%2FK.mol%29%2B2mole%5Ctimes%20%28248J%2FK.mol%29%7D%5D-%5B2mole%5Ctimes%20%28206J%2FK.mol%29%2B3mole%5Ctimes%20%28205J%2FK.mol%29%5D)
Now we have to calculate the Gibbs free energy of reaction 
.
As we know that,
At room temperature, the temperature is 500 K.
Therefore, the value of for the reaction is -959.1 kJ
