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vichka [17]
3 years ago
5

Two objects are brought into contact Object 1 has mass 0.76 kg, specific heat capacity 0.87) g'c and initial temperature 52.2 'C

. Object #2 has mass 70.7 9. specific heat capacity 3.071" "c' and initial temperature 154 *C. What is the final temperature of the two masses after thermal equilibrium has been reached? Assume the two objects are thermally isolated from everything else. Express your answer in C​
Chemistry
1 answer:
taurus [48]3 years ago
4 0

Answer:

T_F=77.4\°C

Explanation:

Hello there!

In this case, according to the given information, it turns out possible to set up the following energy equation for both objects 1 and 2:

Q_1=-Q_2

In terms of mass, specific heat and temperature change is:

m_1C_1(T_F-T_1)=-m_2C_2(T_F-T_2)

Now, solve for the final temperature, as follows:

T_F=\frac{m_1C_1T_1+m_2C_2T_2}{m_1C_1+m_2C_2}

Then, plug in the masses, specific heat and temperatures to obtain:

T_F=\frac{760g*0.87\frac{J}{g\°C} *52.2\°C+70.7g*3.071\frac{J}{g\°C}*154\°C}{760g*0.87\frac{J}{g\°C} +70.7g*3.071\frac{J}{g\°C}} \\\\T_F=77.4\°C

Yet, the values do not seem to have been given correctly in the problem, so it'll be convenient for you to recheck them.

Regards!

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A solution contains 0.036 M Cu2+ and 0.044 M Fe2+. A solution containing sulfide ions is added to selectively precipitate one of
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The precipitate is CuS.

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Explanation:

<u>Step 1: </u>Data given

The solution contains 0.036 M Cu2+ and 0.044 M Fe2+

Ksp (CuS) = 1.3 × 10-36

Ksp (FeS) = 6.3 × 10-18

Step 2:  Calculate precipitate

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[S2-]= 3.61*10^-35 M

<u>For Iron</u>  we have: Ksp=[Fe2+]*[S2-]

Ksp(FeS) = 6.3*10^-18 = 0.044M*[S2-]

[S2-]= 1.43*10^-16 M

CuS will form precipitates before FeS., because only 3.61*10^-35 M Sulfur Ions are needed for CuS. For FeS we need 1.43*10^-16 M Sulfur Ions which is much larger.

The precipitate is CuS.

Sulfide will precipitate at  [S2-]= 3.61*10^-35 M

3 0
3 years ago
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