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3 years ago

During chemical weathering, forsterite is dissolved by the carbonic acid in rainwater. The weathering reaction is as follows:

Chemistry

1 answer:

Semenov [28]3 years ago

4 0

Answer:

Keqq = 310

Note: Some parts of the question were missing. The missing values are used in the explanation below.

Explanation:

<em>Given values: ΔH° = -178.8 kJ/mol = -178800 J/mol; T = 25°C = 298.15 K; ΔS° = -552 J/mol.K; R = 8.3145 J/mol.K</em>

Using the formula ΔG° = -RT㏑Keq

㏑Keq = ΔG°/(-RT)

where ΔG° = ΔH° - TΔS°

㏑Keq = ΔH° - TΔS°/(-RT)

㏑Keq = {-178800 - (-552 * 298.15)} / -(8.3145 * 298.15)

㏑Keq = -14221.2/-2478.968175

㏑Keq = 5.73674166

Keq = e⁵°⁷³⁶⁷⁴¹⁶⁶

Keq = 310.05

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orbital cellulitis 5462 children included for study, 1347 (24%) were prescribed corticosteroids within 2 days of admission

son4ous [18]

It's unable to identify a decrease in LOS linked to corticosteroid exposure during hospitalization for ocular cellulitis in this database search. After two days of hospitalization, operational episodes and the prescription of corticosteroids were related to admission to the PICU.

Within two days of admission, 1347 (24%) of the 5462 children who were included in the research received a corticosteroid prescription. In analyses that controlled for age, the existence of meningitis, abscess, or visual problems, as well as the surgical episode and PICU admission within 2 days, corticosteroid prescription was not linked with LOS (e = 1.01, 95% confidence interval [CI]: 0.97-1.06). Among patients with a primary diagnosis of orbital cellulitis, corticosteroid exposure was linked to surgical events after two days of hospitalization (odds ratio = 2.05, 95% CI: 1.29-3.27) and 30-day readmission (odds ratio = 2.40, 95% CI: 1.52-3.78). Prospective, randomized control trials are required prior to the widespread usage of corticosteroids.

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4 0

2 years ago

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

3 0

3 years ago

How many protons are in an element with an atomic number of 8 and a mass of 18?

Pachacha [2.7K]

B: 8 the amount of protons is equal to the atomic number

4 0

3 years ago

Read 2 more answers

Which has the reverse chemical reaction as

Mazyrski [523]

Answer:

please further explain your question

4 0

3 years ago

Calcium nitrate and ammonium fluoride react to form calcium fluoride, dinitrogen monoxide, and water vapor. How many grams of di

pychu [463]

Answer:

16.79 grams of dinitrogen monoxide are present after 31.3 g of calcium nitrate and 38.7 g of ammonium fluoride react completely. And calcium nitrate is a limiting reactant. Explanation:

$Ca(NO_3)_2(aq)+2NH_4F(aq)\rightarrow CaF_2(aq)+2N_2O(g)+4H_2O(g)$

Moles of calcium nitrate = $\frac{31.3 g}{164 g/mol}=0.1908 mol$

Moles of ammonium fluoride = $\frac{38.7 g}{37 g/mol}=1.046 mol$

According to reaction , 2 moles of ammonium fluoride reacts with 1 mole of calcium nitrate.

Then 1.046 moles of ammonium fluoride will react with :

$\frac{1}{2}\times 1.046 mol=0.523 mol$ calcium nitarte .

This means that ammonium fluoride is in excess amount and calcium nitrate is in limiting amount.

Hence, calcium nitrate is a limiting reactant.

So, amount of dinitrogen monoxide will depend upon moles of calcium nitrate.

So, according to reaction , 1 mole of calcium nitarte gives 2 moles of dinitrogen monoxide gas .

Then 0.1908 moles of calcium nitrate will give:

$\frac{2}{1}\times 0.1908 mol=0.3816 mol$ of dinitrogen monoxide gas.

Mass of 0.03816 moles of dinitrogen monoxide gas:

0.03816 mol × 44 g/mol = 16.79 g

16.79 grams of dinitrogen monoxide are present after 31.3 g of calcium nitrate and 38.7 g of ammonium fluoride react completely. And calcium nitrate is a limiting reactant.

8 0

3 years ago