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1 month ago

Enter the chemical equation 2h (aq) s2−(aq)→h2s(g).

1 answer:

5 0

I’m unsure about what (g) refers to.
It could be that the answer is 2H(aq)S₂−2(aq) ⇒ H₂S₂.

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Calculate ΔH and ΔStot when two copper blocks, each of mass 10.0 kg, one at 100°C and the other at 0°C, are placed in contact in

eduard [2782]

Clarification:

The pertinent information is outlined as follows.

m = 10.0 kg = 10,000 g (since 1 kg = 1000 g)

Starting temperature of block 1, $T_{1} = 100^{o}C$ = (100 + 273) K = 373 K

Starting temperature of block 2, $T_{2} = 0^{o}C$ = (0 + 273) K = 273 K

Therefore, the heat lost by block 1 equals the heat received by block 2

$mC \Delta T = mC \times \Delta T$

$10000 g \times 0.385 \times (T_{f} - 100)^{o}C = 10000 g \times 0.385 \times (0 - T_{f})^{o}C$

$T_{f} - 100^{o}C = 0^{o}C - T_{f}$

$2T_{f} = 100^{o}C$

$T_{f} = 50^{o}C$

It's important to convert the temperature into Kelvin as (50 + 273) K = 323 K.

Additionally, the relationship between enthalpy and temperature change is as follows.

$\Delta H = mC \Delta T$

= $10000 g \times 0.385 J/K g \times 323 K$

= 1243550 J

or, = 1243.5 kJ

Next, determine the entropy change for block 1 as follows.

$\Delta S_{1} = mC ln \frac{T_{f}}{T_{i}}$

= $10000 g \times 0.385 J/K g \times ln \frac{323}{373}$

= $10000 g \times 0.385 J/K g \times -0.143$

= -554.12 J/K

Now, the entropy change for block 2 is as follows.

$\Delta S_{2} = mC ln \frac{T_{f}}{T_{i}}$

= $10000 g \times 0.385 J/K g \times ln \frac{323}{273}$

= $10000 g \times 0.385 J/K g \times 0.168$

= 647.49 J/K

Thus, the total entropy is the sum of the entropy changes of both blocks.

= -554.12 J/K + 647.49 J/K $\Delta S_{total} = \Delta S_{1} + \Delta S_{2}$

= 93.37 J/K

In conclusion, for this reaction, the outcome is 1243.5 kJ and $\Delta S_{total}$ is 93.37 J/K.

6 0

28 days ago

Determine how many grams of silver would be produced, if 12.83 x 10^23 atoms of copper react with an excess of silver nitrate. G

Anarel [2989]

1) The chemical equation is

Cu + 2AgNO3 ---> Cu (NO3)2 + 2Ag

2) Molar ratios are as follows:

1 mol Cu: 2 moles AgNO3: 1 mol Cu (NO3)2: 2 mol Ag

3) Converting 12.83 * 10^23 atoms of Cu to moles gives:

12.83 * 10^23 atoms / (6.02 * 10^23 atoms / mol) = 2.131 mol Cu

4) Using the ratios:

2.131 mol Cu * 2 mol Ag / 1 mol Cu = 4.262 mol Ag

5) To convert 4.262 mol of silver to grams, use the atomic weight of silver:

mass = moles × atomic mass = 4.262 mol * 107.9 g / mol = 459.9 grams

Answer: 459.9 g

5 0

1 month ago

(a) calculate the %ic of the interatomic bond for the intermetallic compound tial3. (b) on the basis of this result, what type o

Tems11 [2777]

Answer :

The percentage ionic character (%IC) equals 10%, indicating the bond is mostly covalent with slight polarity.

Percent Ionic Character:

This reflects the fraction of ionic nature within a polar covalent bond. The formula for %IC (% ionic character) is:

$Percent Ionic character = 1 - e^-^0^.^2^5 ^*^(^X^a^-^X^b^) * 100$