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Setler79 [48]
4 years ago
6

\The specific heat of aluminum is 0.21 cal g°C . How much heat is released when a 10 gram piece of aluminum foil is taken out of

the oven and cools from 50 to 10 °C?
Chemistry
1 answer:
atroni [7]4 years ago
8 0

<u>Answer: </u>The amount of heat released is 84 calories.

<u>Explanation: </u>

The equation used to calculate the amount of heat released or absorbed, we use the equation:

Q= m\times c\times \Delta T

where,

Q = heat gained  or released = ? Cal

m = mass of the substance = 10g

c = specific heat of aluminium = 0.21 Cal/g ° C

Putting values in above equation, we get:

\Delta T={\text{Change in temperature}}=(10-50)^oC=-40^oC  

Q=10g\times 0.21Cal/g^oC\times-40^oC

Q = -84 Calories

Hence, the amount of heat released is 84 calories.

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After performing a dilution calculation, you determine you need 25.0 milliliters of an aqueous stock solution to make 100.0 mill
IRINA_888 [86]
A is obviously out because it leads to a volume of 125.0 milliliters of the new solution and gives you a lower concentration than you were aiming for.

D is out because you are adding 75 milliliters of the stock solution, so your concentration would be too high. You only need 25.0 milometers of stock solution per 100 milliliters of the new solution.

C is also out because it leads to 50.0 milliliters stock solution per 100 milliliters of the new solution and hence the wrong concentration.

B is by default the correct answer. It also details the correct technique. First you add the stock solution (This you know from your calculations to be 25 milliliters.) then you add the water up to the volume you needed. (Because the calculations only tell you the total volume of water not what you need to add) You also add the water last so you can rinse the neck of the flask to make sure you also get all the stock solution residue into the stock solution.

I would add the final step of stirring, but B is the only answer that can be correct.
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4 years ago
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Answer:

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Explanation:

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6 0
3 years ago
A 36.2 g object has a heat capacity of 12.5 J/ oC. How much energy (in J) is required to raise the temperature by 5.73 oC?
iren2701 [21]

Answer:

\Delta H=2592.825\ J

Explanation:

The expression for the calculation of the enthalpy change of a process is shown below as:-

\Delta H=m\times C\times \Delta T

Where,  

\Delta H  is the enthalpy change

m is the mass

C is the specific heat capacity

\Delta T  is the temperature change

Thus, given that:-

Mass of object = 36.2 g

Specific heat = 12.5 J/g°C

\Delta T=5.73\ ^0C

So,  

\Delta H=36.2\times 12.5\times 5.73\ J=2592.825\ J

\Delta H=2592.825\ J

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Evaluate the exponential expression (−2)6.
Olegator [25]

A general exponential expression is something like:

A^n

This means that we need to multiply the number A by itself n times.

Using that we will get (-2)^6 = 64

With that definition, we can rewrite:

(-2)^6 = (-2)*(-2)*(-2)*(-2)*(-2)*(-2)

So we just need to solve the above expression.

Also, remember the rule of signs:

(-)*(-) = (+)

We will get:

(-2)*(-2)*(-2)*(-2)*(-2)*(-2) =  [(-2)*(-2)]*[(-2)*(-2)]*[(-2)*(-2)]

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Then we got:

(-2)^6 = 64

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