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Natasha2012 [34]
2 years ago
7

If a piece of cadmium with a mass of 37.60 g and a temperature of 100.0 oC is dropped into 25.00 cc of water at 23.0 oC, what wi

ll be the final temperature of the system
Chemistry
1 answer:
zlopas [31]2 years ago
6 0

Answer:

T_{eq}=28.9\°C

Explanation:

Hello!

In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:

Q_{Cd}+Q_{W}=0

Thus, we insert mass, specific heat and temperatures to obtain:

m_{Cd}C_{Cd}(T_{eq}-T_{Cd})+m_{W}C_{W}(T_{eq}-T_{W})=0

In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:

T_{eq}=\frac{m_{Cd}C_{Cd}T_{Cd}+m_{W}C_{W}T_{W}}{m_{Cd}C_{Cd}+m_{W}C_{W}}

Now, we plug in to obtain:

T_{eq}=\frac{37.60g*0.232\frac{J}{g\°C}*100.00\°C+25.00g*4.184\frac{J}{g\°C}*23.0\°C}{37.60g*0.232\frac{J}{g\°C}+25.00g*4.184\frac{J}{g\°C}}\\\\T_{eq}=28.9\°C

NOTE: since the density of water is 1g/cc, we infer that 25.00 cc equals 25.00 g.

Best regards!

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Linda performed the following trials in an experiment. Trial 1: Heat 30.0 grams of water at 0 °C to a final temperature of 40.0
nexus9112 [7]

<u>Answer:</u> The correct answer is Option b.

<u>Explanation:</u>

To calculate the amount of heat absorbed or released, we use the following equation:

q=mc\Delta T    .....(1)

where, q = amount of heat absorbed or released.

m = mass of the substance

c = heat capacity of  water = 4.186 J/g ° C      

\Delta T = Change in temperature

  • <u>For Trial 1:</u>

We are given:

m=30g\\\Delta T=[40-0]^oC=40^oC\\q=?J

Putting values in equation 1, we get:

q=30g\times 4.186J/g^oC\times 40^oC

q = 5023.2 J

  • <u>For trial 2:</u>

We are given:

m=40g\\\Delta T=[40-30]^oC=10^oC\\q=?J

Putting values in equation 1, we get:

q=40g\times 4.186J/g^oC\times 10^oC

q = 1674.4 J

Heat gained by Trial 1 than trial 2 = (5023.2-1674.4)J=3347J

Hence, the amount of heat gained in Trial 1 about 3347 J more than the heat released in Trial 2.

Thus, the correct answer is Option b.

4 0
3 years ago
Use the following data to estimate ΔHf° for magnesium fluoride. Mg(s) + F2(g) → MgF2(s) lattice energy −2913 kJ/mol first ioniza
miss Akunina [59]

Answer:

The correct answer is -1085 KJ/mol

Explanation:

To calculate the formation enthalphy of a compound by knowing its lattice energy, you have to draw the Born-Haber cycle step by step until you obtain each element in its gaseous ions. Find attached the correspondent Born-Haber cycle.

In the cycle, Mg(s) is sublimated (ΔHsub= 150 KJ/mol) to Mg(g) and then atoms are ionizated twice (first ionization: ΔH1PI= 735 KJ/mol, second ionization= 1445 KJ/mol) to give the magnesium ions in gaseous state.

By other hand, the covalent bonds in F₂(g) are broken into 2 F(g) (Edis= 154 KJ/mol) and then they are ionizated to give the fluor ions in gaseous state 2 F⁻(g) (2 x ΔHafinity=-328 KJ/mol). The ions together form the solid by lattice energy (ΔElat=-2913 KJ/mol).

The formation enthalphy of MgF₂ is:

ΔHºf= ΔHsub + Edis + ΔH1PI + ΔH2PI + (2 x ΔHaffinity) + ΔElat

ΔHºf= 150 KJ/mol + 154 KJ/mol + 735 KJ/mol + 1445 KJ/mol + (2 x (-328 KJ/mol) + (-2913 KJ/mol).

ΔHºf= -1085 KJ/mol

3 0
3 years ago
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