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3 years ago

Theres Only 1 Thing

Chemistry

2 answers:

VARVARA [1.3K]3 years ago

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jews are big fun

explanation:

theyjustare

xz_007 [3.2K]3 years ago

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I love youuuuuuuuuuuuuuuuyyy!!!!!!!!!!!!!!!

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Giving brainly if correct

Kamila [148]

Answer:

Explanation:

filtration, the process in which solid particles in a liquid or gaseous fluid are removed by the use of a filter medium that permits the fluid to pass through but retains the solid particles.

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Can I have brain pls

6 0

2 years ago

Read 2 more answers

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

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

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

Now put all the given values in this expression, we get:

$\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]$

$\Delta S^o=-153J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 500 K.

$\Delta G^o=(-1035600J)-(500K\times -153J/K)$

$\Delta G^o=-959100J=-959.1kJ$

Therefore, the value of $\Delta G^o$ for the reaction is -959.1 kJ

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3 years ago

Identify the reactants and the products of the preparation step to the urea cycle.

Salsk061 [2.6K]

According to sources, the most probable answer to this query is the enzymes and waste products that are collected by the nephron from the blood. Thank you for your question. Please don't hesitate to ask in Brainly your queries.

7 0

3 years ago

A chemical engineer is developing a process for producing a new chemical. One step in the process involves allowing a solution o

notka56 [123]

increasing temperature

8 0

4 years ago

Read 2 more answers

Reaction 1: C3H8(g)+5O2(g)→3CO2(g)+4H2O(g), ΔH1=−2043 kJ

Leokris [45]

Given teh equation adn the heat of reaction, reaction 2's heat of reaction can be obtained by simply multiplying teh heat of reaction of 1 by 3. The final answer is -6129 kJ.

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3 years ago