Answer:
THE SPECIFIC HEAT OF THE ALLOY IS 0.9765 J/g K
Explanation:
Mass of alloy = 33 g
Initial temperature of alloy = 93°C
Mass of water = 50 g
Initail temp. of water = 22 °C
Heat capacity of calorimeter = 9.20 J/K
Final temp. = 31.10 °C
specific heat of alloy = unknown
specific heat capacity of water = 4.2 J/g K
Heat = mass * specific heat * change in temperature = m c ΔT
Heat = heat capcity * chage in temperature = Δ H * ΔT
In calorimetry;
Heat lost by the alloy = Heat gained by water + Heat of the calorimeter
mc ΔT = mcΔT + Heat capacity * ΔT
33 * C * ( 93 - 31.10) = 50 * 4.2 * ( 31.10 -22) + 9.20 * ( 31.10 -22)
33 * C * 61.9 = 50 * 4.2 * 9.1 + 9.20 * 9.1
2042.7 C = 1911 + 83,72
C = 1911 + 83.72 / 2042.7
C = 1994.72 /2042.7
C =0.9765 J/g K
The specific heat of the alloy is 0.9765 J/ g K
Answer:
molarity of acid =0.0132 M
Explanation:
We are considering that the unknown acid is monoprotic. Let the acid is HA.
The reaction between NaOH and acid will be:
Thus one mole of acid will react with one mole of base.
The moles of base reacted = molarity of NaOH X volume of NaOH
The volume of NaOH used = Final burette reading - Initial reading
Volume of NaOH used = 22.50-0.55= 21.95 mL
Moles of NaOH = 0.1517X21.95=3.33 mmole
The moles of acid reacted = 3.33 mmole
The molarity of acid will be = 
I will list them from alkaline with the lowest boiling point and alkaline with the highest.
1. C2H6
2. C9H20
3. C11H24
4. C16H34
5. C20H42
6. C32H66
7. C150H302
I have taken a quiz similar to this before and can assure you this is correct and is primarily because of the number of Carbons and Hydrogens within this. More Carbons and Hydrogens causes Boiling Points to increase because of stronger bonds.
