Should industries releasing heavy metals into land and water ecosystem be penalized

A calorimeter contains 72.0 g of water at 19.2 oC. A 141 g piece of metal is heated to 89.0 oC and dropped into the water. The e

kumpel [21]

Answer:

The specific heat of the metal is 0.212 J/(g°C).

Explanation:

We can calculate the specific heat of the metal by the following equilibrium:

$q_{a} = -q_{b}$

$m_{a}C_{a}\Delta T_{a} = -m_{b}C_{b}\Delta T_{b}$

$m_{a}C_{a}(T_{f_{a}} - T_{i_{a}}) = -m_{b}C_{b}(T_{f_{b}} - T_{i_{b}})$

In the above equation, we have that the heat loses by the metal (b) is gained by the water (a).

$m_{a}$ : is the water's mass = 72.0 g

$C_{a}$ : is the specific heat of water = 4.184 J/(g°C)

$T_{i_{a}}$ : is the initial temperature of the water = 19.2 °C

$T_{f_{a}}$ : is the final temperature of the water = 25.5 °C

$m_{b}$ : is the metal's mass = 141 g

$C_{b}$ : is the specific heat of metal =?

$T_{i_{b}}$ : is the initial temperature of the metal = 89.0 °C

$T_{f_{b}}$ : is the final temperature of the water = 25.5 °C

$m_{a}C_{a}(T_{f_{a}} - T_{i_{a}}) = -m_{b}C_{b}(T_{f_{b}} - T_{i_{b}})$

$72.0 g*4.184 J/(g^{\circ} C)*(25.5 ^{\circ} C - 19.2 ^{\circ} C) = -141 g*C_{b}*(25.5 ^{\circ} C - 89.0 ^{\circ} C)$

$C_{b} = -\frac{72.0 g*4.184 J/(g^{\circ} C)(25.5 ^{\circ} C - 19.2 ^{\circ} C)}{141 g(25.5 ^{\circ} C - 89.0 ^{\circ} C)} = 0.212 J/(g^{\circ} C)$

Therefore, the specific heat of the metal is 0.212 J/(g°C).

I hope it helps you!

7 0

3 years ago

Noble gas compounds like KrF, XeCl, and XeBr are used in excimer lasers. Draw an approximate molecular orbital diagram appropria

Aneli [31]

Answer:

Here's what I get.

Explanation:

The MO diagrams of KrBr, XeCl, and XeBr are shown below.

They are similar, except for the numbering of the valence shell orbitals.

Also, I have drawn the s and p orbitals at the same energy levels for both atoms in the compounds. That is obviously not the case.

However, the MO diagrams are approximately correct.

The ground state electron configuration of KrF is

$(1\sigma_{g})^{2}\, (1\sigma_{u}^{*})^{2} \, (2\sigma_{g})^{2} \, (2\sigma_{u}^{*})^{2} \, (3\sigma_{g})^{2} \, (1\pi_{u})^{4} \, (1\pi_{g}^{*})^{4} \, (3\sigma_{g}^{*})^{1}$

KrF⁺ will have one less electron than KrF.

You remove the antibonding electron from the highest energy orbital, so the bond order increases.

The KrF bond will be stronger.

6 0

4 years ago

What defines the group of organisms referred to as protist

avanturin [10]

Well protists quite very as organisms i.e it is hard to say if an organism is a protist. Some main features of protists are that they have
1. A membrane enclosed nucleus
2.They are mainly single celled

Also protista are found mainly anywhere water is present for example damp soil, pools and lakes.

Hope this helps :).

3 0

4 years ago

Who discovered cells come from other cells

Natalija [7]

Answer: Rudolf Virchow

3 0

3 years ago

What kind of energy is stored in an electromagnet?

meriva

Kinetic is the answer to your question

6 0

3 years ago