At the melting point. Draw a line up from 0 degrees and a line to the right from 1 atm. They meet at the line between solid and liquid... the melting point
Question is incomplete, complete question is;
A 34.8 mL solution of 
(aq) of an unknown concentration was titrated with 0.15 M of NaOH(aq).
If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of 
? For your answer, only type in the numerical value with two significant figures. Do NOT include the unit.
Answer:
0.044 M is the molarity of (aq).
Explanation:
The reaction taking place here is in between acid and base which means that it is a neutralization reaction .
To calculate the concentration of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is
are the n-factor, molarity and volume of base which is NaOH.
We are given:
Putting values in above equation, we get:
0.044 M is the molarity of (aq).
Co= 59g
S=32g
59+32=91g/mol
Note: The question is incomplete. The complete question is given below :
Suppose a substance has a heat of fusion equal to 45 cal/g and a specific heat of 0.75 cal/g°C in the liquid state. If 5.0 kcal of heat are applied to a 50 g sample of the substance at a temperature of 24°C, what will its new temperate be? What state will the sample be in? (melting point of the substance = 27°C; specific heat of the solid =0.48 cal/g°C; boiling point of the substance = 700°C)
Explanation:
1.a) Heat energy required to raise the temperature of the substance to its melting point, H = mcΔT
Mass of solid sample = 50 g; specific heat of solid = 0.75 cal/g; ΔT = 27 - 24 = 3 °C
H = 50 × 0.75 × 3 = 112.5 calories
b) Heat energy required to convert the solid to liquid at its melting point at 27°C, H = m×l, where l = 45 cal/g
H = 50 × 45 = 2250 cal
c) Total energy used so far = 112.5 cal + 2250 cal = 2362.5 calories.
Amount of energy left = 5000 - 2362.5 = 2637.5 cal
The remaining energy is used to heat the liquid
H = mcΔT
Where specific heat of the liquid, c = 0.75 cal/g/°C, H = 2637.5 cal, ΔT = temperature change
2637.5 = 50 × 0.75 x ΔT
ΔT = 2637.5 / ( 50*0.75)
ΔT = 70.3 °C
Final temperature of sample = (70.3 + 27) °C = 97.3 °C
The substance will be in liquid state at a temperature of 97.3 °C
i hope that this eg gonna help u
