Chemistry... please help me

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Ответ:

NorbxrtThaG

06.11.2022

Chemistry

Aluminum

Explanation:

Given

T_1 = 25.2^oC

T_2 = 55.1^oC

m = 5.00g

$\triangle Q= 133J$

See attachment for chart

Required

Identify the unknown substance

To do this, we simply calculate the specific heat capacity from the given parameters using:

$c = \frac{\triangle Q}{m\triangle T}$

This gives:

$c = \frac{\triangle Q}{m(T_2 - T_1)}$

So, we have:

$c = \frac{133J}{5.00g * (55.1C - 25.2C)}$

$c = \frac{133J}{5.00g * 29.9C}$

$c = \frac{133J}{149.5gC}$

$c = 0.89\ J/gC$

From the attached chart, we have:

$Al(s) = 0.89\ J/gC$ --- The specific heat capacity of Aluminum

Hence, the unknown substance is Aluminum

5. A 5.00 g sample of an unknown substance was heated from 25.2 C to 55.1 degrees * C , and it requi

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