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2 years ago

Question 4 (1 point)

Chemistry

1 answer:

MArishka [77]2 years ago

7 0

The tongue depressor will float on water because its density is less than that of water.

<h3>When do substances sink or float in a liquid?</h3>

Substance will either float or sink in a liquid depending on the comparative densities of the substance and the liquid.

A substance will float in a liquid if it is less dense than the liquid.

A substance will sink in a liquid if it is denser than the liquid.

The density of the tongue depressor = 4.54/6.78 = 0.67 g/cm³

The density of water is 1 g/cm³.

Therefore, the tongue depressor will float on water.

In conclusion, the density of substances determines whether thy float or sink in water.

Learn more about density at: brainly.com/question/952755

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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

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<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

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3 years ago

Which must be the same when comparing 1 mol of oxygen, O2, with 1 mol of carbon monoxide has, CO?

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Answer: 1 mol of oxygen, O₂, and 1 mol of CO will have the same number of molecules, and the same number of atoms.

Justification:

Althought the question is too open, other answers may arise, the most remarkable similarity between the two compounds is that both are diatomic.

That means that both molecules oxygen, O₂, and carbon monoxide, CO have two atoms.

So, 1 mol of oxygen, O₂, and 1 mol of CO will have the same number of molecules, and the same number of atoms.

You must remember that 1 mol means a specific number. It is Avogadro's number, which is 6.022 × 10 ²³.

So 1 mol of CO and 1 mol of O₂ are the same number of representative particles: 6.022 ×10²³ molecules eac, and two times that number of atoms each (since each molecule has two atoms).

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