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3 years ago

Explain why, at room temperature, CO2 is a gas and CS2 is a liquid. You may use diagrams as part of your explanation.

Chemistry

1 answer:

eduard3 years ago

3 0

Explanation:

A. Hydrogen bonding is present in CS2 but not in CO2.

B. CS2 has greater dipole moment than CO2 and thus the dipole-dipole forces in CS2 are stronger.

C. CS2 partly dissociates to form ions and CO2 does not. Therefore, ion-dipole interactions are present in CS2 but not in CO2.

D. The dispersion forces are greater in CS2 than in CO2.

PLS MARK BRAINLIEST :D

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3 years ago

If you have 110.0 grams of an unknown compound that contains 12.3 grams of hydrogen, what is the percent by mass of hydrogen in

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For the reaction Fe3O4(s) + 4H2(g) --> 3Fe(s) + 4H2O(g)

mojhsa [17]

Answer : The value of equilibrium constant for this reaction at 328.0 K is $1.70\times 10^{15}$

Explanation :

As we know that,

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

where,

$\Delta G^o$ = standard Gibbs free energy = ?

$\Delta H^o$ = standard enthalpy = 151.2 kJ = 151200 J

$\Delta S^o$ = standard entropy = 169.4 J/K

T = temperature of reaction = 328.0 K

Now put all the given values in the above formula, we get:

$\Delta G^o=(151200J)-(328.0K\times 169.4J/K)$

$\Delta G^o=95636.8J=95.6kJ$

The relation between the equilibrium constant and standard Gibbs free energy is:

$\Delta G^o=-RT\times \ln k$

where,

$\Delta G^o$ = standard Gibbs free energy = 95636.8 J

R = gas constant = 8.314 J/K.mol

T = temperature = 328.0 K

K = equilibrium constant = ?

Now put all the given values in the above formula, we get:

$95636.8J=-(8.314J/K.mol)\times (328.0K)\times \ln k$

$k=1.70\times 10^{15}$

Therefore, the value of equilibrium constant for this reaction at 328.0 K is $1.70\times 10^{15}$

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3 years ago

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4 years ago

when 10.00 g of phosphorus reacts with oxygen, it produces 17.77 g of a phosphorus oxide. This phosphorus oxide was found to hav

Alex_Xolod [135]

Answer:

Molecular formula = P₄O₆

Explanation:

P(s) + O₂(g)------------------------------------⇒ PₓOₙ (g)

10g (17.77-10)g 17.77g

10g 7.77g 17.77g (gramme ratio)

The molecular mass of Phosphorus (P) = 31g/mole

The molecular mass of Oxygen atom (O) = 16g/mole

Mole ratio is given by:

P : O

10/31 7.77/16

0.3226 : 0.4856 Mole ratio---------------------------- (1)

Divide (1) through by 0.3226

1 : 1.5-------------------------------------------- (2)

From (2), the empirical formula for Phosphorus oxide :

Empirical formula = P₁O₁.₅

= PO₁.₅

The molecular formula can be calculated from below:

Since the molecular formula is a multiple of the empirical formula we have

Molecular formula = (PO₁.₅)ₙ----------------------------------- (3)

Since we are given the molecular mass of the oxide formed, we have:

(PO₁.₅)ₙ = 220-----------------------------(4)

[31 + (16 x 1.5)] x n = 220

[31 + 24]n = 220

55n =220

n = 4

Substituting into (3), we have :

Molecular formula = (PO₁.₅)₄

= P₄O₆

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