Energy and specific heat

1. draw a graph of an exothermic reaction. label reactants, products and ∆h.

2. calculate the amount of energy required to raise the temperature of 3.00g of gold from 45.9 to 93.0°c.

3. 1.70g of a silvery metal requires 1000.j of energy to change its temp from 298k to 2749k. is the metal pure silver?

1. Kindly, see the attached image.

2. 18.369 J.

3. Yes, the metal is pure silver.

Explanation:

1. Draw a graph of an exothermic reaction. Label reactants, products and ∆H.

2. Calculate the amount of energy required to raise the temperature of 3.00 g of gold from 45.9 to 93.0°C.

Q = m.c.ΔT.

Where, Q is the amount of energy required to raise the temperature of gold (J).

m is the mass of the gold (m = 3.0 g).

c is the specific heat of gold (c = 0.13 J/g.°C).

ΔT is the temperature difference between the initial and final T (ΔT = 93.0 °C - 45.9 °C = 47.1 °C).

∴ Q = m.c.ΔT = (3.0 g)(0.13 J/g.°C)(47.1 °C) = 18.369 J.

3. 1.70 g of a silvery metal requires 1000.0 J of energy to change its temp from 298 K to 2749 K. Is the metal pure silver?

As mentioned in details in the second point, we can use the relation:

Q = m.c.ΔT

Q = 1000.0 J, m = 1.70 g, ΔT = 2749 - 298 = 2451 °C.

∴ c = Q/(m.ΔT) = (1000.0 J)/(1.70 g)(2451 °C) = 0.239 J/g.°C ≅ 0.24 J/g.°C.

  the   balanced   equation   for   reaction of   ca metal   with   o2 to   produce   cao   is as below

2 ca+ o2→ 2 cao2

  explanation

the reaction is   balanced     since the   number of atoms   in reactant side is   equal to   number of atoms   in   product side.   for   example   they   are 2   atom of   ca in   reactant   side   and   2   atoms of ca   in   the product side.

  the   process   is a   meet the   requirement   of   redox   reaction because :     calcium   is   oxidized   from   oxidation state 0   to 2+   while   oxygen   is reduced from   oxidation   state   0   to 2-

the missing options are :

(1) lower boiling point and a lower freezing point(2) lower boiling point and a higher freezing point(3) higher boiling point and a lower freezing point(4) higher boiling point and a higher freezing point

answer : the correct answer is option (3) higher boiling point and a lower freezing point

explanation :

when a non volatile solute is added to a solvent like water, it results in elevation of boiling point.

let us find the increase in boiling point of water when cacl₂ is added to water to make a solution of 1 m

step 1 : find molality of the solution

let us assume we have 1 l of solution.

let us convert this to grams of cacl₂.

molar mass of cacl2 is 111 g/mol

grams of cacl₂ =

mass of solute = 111 g

assuming density of solution same as water which is 1g/ml we have ,

mass of solution = density * volume

mass of solution =

mass of solution = mass of solvent + mass of solute

mass of solute = 1000g - 111g

mass of solute 889 g = 0.889 kg

the molality of the solution can be calculated as

molality = 1.12 m

step 2 : find boiling point elevation.

the elevation in boiling point is calculated as

here δtb = boiling point elevation

i = vant hoff factor which is the number of ions formed by the solute .

cacl₂ dissociates as

it forms 3 ions . therefore,

i = 3

m = molality which is 1.12 m

kb is constant for a given solvent. for water it is 0.512

the boiling point of water increases by 1.7 °c

step 3: find freezing point depression.

when a non volatile solute is added to water, it results in decrease in freezing point.

the formula to find freezing point depression is

kf is constant for water and its value is 1.86

the freezing point of water decreases by 6.2 °c

therefore we can say that, the solution will have a higher boiling point and a lower freezing point