Determine the volume at STP of 36.8 g of nitrogen dioxide(NO2)

A sample of iron is put into a calorimeter (see sketch at right) that contains of water. The iron sample starts off at and the t

Mariulka [41]

Answer:

Therefore, the specific heat capacity of the iron is 0.567J/g.°C.

<em>Note: The question is incomplete. The complete question is given as follows:</em>

<em>A 59.1 g sample of iron is put into a calorimeter (see sketch attached) that contains 100.0 g of water. The iron sample starts off at 85.0 °C and the temperature of the water starts off at 23.0 °C. When the temperature of the water stops changing it's 27.6 °C. The pressure remains constant at 1 atm. </em>

<em> Calculate the specific heat capacity of iron according to this experiment. Be sure your answer is rounded to the correct number of significant digits</em>

Explanation:

Using the formula of heat, Q = mc∆T

where Q = heat energy (Joules, J), m = mass of a substance (g)

c = specific heat capacity (J/g∙°C), ∆T = change in temperature (°C)

When the hot iron is placed in the water, the temperature of the iron and water attains equilibrium when the temperature stops changing at 27.6 °C. Since it is assumed that heat exchange occurs only between the iron metal and water; Heat lost by Iron = Heat gained by water

mass of iron = 59.1 g, c = ?, Tinitial = 85.0 °C, Tfinal = 27.6 °C

∆T = 85.0 °C - 27.6 °C = 57.4 °C

mass of water = 100.0 g, c = 4.184 J/g∙°C, Tinitial = 23.0 °C, Tfinal = 27.6 °C

∆T = 27.6°C - 23.0°C = 4.6 °C

Substituting the values above in the equation; Heat lost by Iron = Heat gained by water

59.1 g * c * 57.4 °C = 100.0 g * 4.184 J/g.°C * 4.6 °C

c = 0.567 J/g.°C

Therefore, the specific heat capacity of the iron is 0.567 J/g.°C.

5 0

3 years ago

The Loch Ness monster is a sea serpent that is thought to live in a lake in

Dmitry_Shevchenko [17]

Answer:

Scotland

Explanation:

5 0

3 years ago

Read 2 more answers

Water has a specific heat of 4.186 J/g*C. How much energy would be required to raise the temperature of 10 g of water by 10 C?

Sladkaya [172]

Energy(heat) required to raise the temperature of water : 418.6 J

<h3>Further explanation </h3>

Heat can be calculated using the formula:

Q = mc∆T

Q = heat, J

m = mass, g

c = specific heat, joules / g ° C

∆T = temperature difference, ° C / K

Specific heat of water = 4.186 J/g*C.

∆T(raise the temperature) : 10° C

mass = 10 g

Heat required :

$\tt Q=m.c.\Delta T\\\\Q=10\times 4.186\times 10\\\\Q=418.6~J$

8 0

3 years ago

Plz help!!! This is timed!!!

pogonyaev

58.7 %

Please correct me if I’m wrong. :)

7 0

2 years ago

The equilibrium constant has been estimated to be 0.12 at 25 °C. If you had originally placed 0.069 mol of cyclohexane in a 2.8

scZoUnD [109]

Answer: Concentrations of cyclohexane and methylcyclopentane at equilibrium are 0.0223 M and 0.0027 M respectively

Explanation:

Moles of cyclohexane = 0.069 mole

Volume of solution = 2.8 L

Initial concentration of cyclohexane = $\frac{moles}{Volume}=\frac{0.069}{2.8}=0.025M$

The given balanced equilibrium reaction is,

cyclohexane ⇔ methylcyclopentane

Initial conc. 0.025 M 0

At eqm. conc. (0.025-x)M (x) M

The expression for equilibrium constant for this reaction will be,

K= methylcyclopentane / cyclohexane

Now put all the given values in this expression, we get :

$0.12=\frac{(x)}{(0.025-x)}$

By solving the term 'x', we get :

x = 0.0027

Concentration of cyclohexane at equilibrium = (0.025-x ) M = (0.025-0.0027) M = 0.0223 M

Concentration of methylcyclopentane at equilibrium = (x ) M = (0.0027) M

4 0

4 years ago