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4 years ago

Q3. A bridge is built without expansion gaps.

Chemistry

2 answers:

goldenfox [79]4 years ago

7 0

Are you learning about atoms?

marshall27 [118]4 years ago

5 0

Answer:

Q3. A bridge is built without expansion gaps.

Explain what could happen to the bridge if the temperature became:

a) much hotter than the day it was built.

b) much colder than the day it was built.

P.s. (this is not a select option question, explain each according to the question)

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Why the iron nail not rusted on a chemical to absorb water vapour

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Thus there was no accumulation of ions on the nails, as the dissolved gases were already removed, hence no rust formation. In test tube C, a chemical was added to absorb the water vapour. We know that rust formation takes place when a metal is exposed to water vapour.

Explanation:i hope u have a wonderful night and stays safe

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3 years ago

_____ sedimentary rocks are made of once-living things or their pieces. A. Organic B. Inorganic C. Chemical D. Fragmental

kramer

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3 years ago

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mel-nik [20]

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3 years ago

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Ann [662]

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2 years ago

Using standard heats of formation, calculate the standard enthalpy change for the following reaction. 2H2S(g) 3O2(g)2H2O(l) 2SO2

Marizza181 [45]

Answer:

$\Delta _rH=-1124.14kJ/mol$

Explanation:

Hello!

In this case, since the standard enthalpy change for a chemical reaction is stood for the enthalpy of reaction, for the given reaction:

$2H_2S(g) +3O_2(g)\rightarrow 2H_2O(l) +2SO_2(g)$

We set up the enthalpy of reaction considering the enthalpy of formation of each species in the reaction at the specified phase and the stoichiometric coefficient:

$\Delta _rH=2\Delta _fH_{H_2O,liq}+2\Delta _fH_{SO_2,gas}-2\Delta _fH_{H_2S,gas}-3\Delta _fH_{O_2,gas}$

In such a way, by using the NIST database, we find that:

$\Delta _fH_{H_2O, liq}=-285.83kJ/mol\\\\\Delta _fH_{SO_2, gas}=-296.84kJ/mol\\\\\Delta _fH_{O_2,gas}=0kJ/mol\\\\\Delta _fH_{H_2S,gas}=-20.50kJ/mol$

Thus, we plug in the enthalpies of formation to obtain:

$\Delta _rH=2(-285.73kJ/mol)+2(-296.84kJ/mol)-2(-20.50kJ/mol)-3(0kJ/mol)\\\\\Delta _rH=-1124.14kJ/mol$

Best regards!

8 0

3 years ago