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3 years ago

Balancing Equations. How many of the element should I drag to the center?

Chemistry

1 answer:

babunello [35]3 years ago

8 0

Answer:

4FeS + 7O₂ ----> 2Fe₂O₃ + 4SO₂

Explanation:

You have to drag the elements shown on the right under the formula. For example, you would have to drag 4 of the FeS molecules under the 4FeS text in the formula. Then place 7 O₂ molecules under the 7O₂ text, etc.

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CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7

padilas [110]

<u>Answer:</u> The $\Delta G$ for the reaction is 54.425 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

$CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)$

We are given:

$\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol$

To calculate $\Delta G^o_{rxn}$ for the reaction, we use the equation:

$\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]$

For the given equation:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J$

Conversion factor used = 1 kJ = 1000 J

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}$

We are given:

$p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm$

Putting values in above equation, we get:

$K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85$

To calculate the gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = $8.314J/K mol$

T = Temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol$

Hence, the $\Delta G$ for the reaction is 54.425 kJ/mol

7 0

3 years ago

What is the ozone hole and how does it affect humans?

denis-greek [22]

Answer:

Ozone layer depletion causes increased UV radiation levels at the Earth's surface, which is damaging to human health. Negative effects include increases in certain types of skin cancers, eye cataracts and immune deficiency disorders.

6 0

3 years ago

Read 2 more answers

Point How many grams of NaCl are present in 11.00 moles?

OlgaM077 [116]

Answer:

642.8g

Explanation:

NaCl has a molar mass of 58.44g/mol, which means one mole of NaCl has a mass of 58.44g.

11 moles of NaCl will have a mass of 642.8g.

11 mol NaCl x 58.44g = 642.8g

6 0

3 years ago

draw the structure of a compound with the formula c9h20 that has two non-adjacent chirality centers, both r.

4vir4ik [10]

Answer:

Explanation:

6 0

2 years ago

Molecular mass of potassium hydrogen carbonate (KHCO3)

adelina 88 [10]

Answer:

100gmol⁻¹

Explanation:

Problem: To calculate the molecular mass of

KHCO₃

Solution

The molar mass of a substance is defined as the number of grams per mole of that substance.

To determine the molar mass of this compound we find:

> the component atomic masses

>we express it in gmol⁻¹

From the periodic table, the atomic masses of the elements in the given compound are:

K = 39g

H = 1g

C = 12g

O = 16g

Now let's add these numbers together:

KHCO₃ = [39 + 1 + 12 + (16x3)]

= 39 +13 +48

= 100g

Molar mass of KHCO₃ = 100gmol⁻¹

7 0

3 years ago