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Nadusha1986 [10]

3 years ago

14

when a sample of a gas is cooled it condenses into a liquid. in which of the following ways do the molecules of the original gas

sample compare with the molecules of a liquid.

1 answer:

Dmitrij [34]3 years ago

4 0

Answer:

The molecules have shorter inter molecular distances and less kinetic energy than the original gas sample.

The cooling reduces the kinetic energy of the gas sample and thus the molecules making up the liquid sample travel less distances due to the less energy. The less inter molecular distances increases the attraction due to inter molecular forces packing the molecules near each other, forming a liquid.

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A student carried out a titration using HC2H3O2(aq)HC2H3O2(aq) and NaOH(aq)NaOH(aq). The net ionic equation for the neutralizati

AnnZ [28]

Answer:

The amount of HC₂H3₃2(aq) in the flask after the addition of 5.0mL of NaOH(aq) compared to the amount of HC₂H₃O₂(aq) in the flask after the addition of 1.0mL is much smaller because more HC₂H₃O₂(aq) is required to react with 5.0 mL NaOH than with 1.0 mL NaOH.

Explanation:

Equation of the reaction between acetic acid, HC₂H₃O₂(aq) and sodium hydroxide, NaOH(aq) is given below:

CH₃COOH (aq) + NaOH (aq) ----> CH₃COONa (aq) + H₂O

The equation of the reaction shows that acetic acid andsodium hydroxide will react in a 1:1 ratio

Since the concentration of NaOH was not given, we can assume that the concentration is 0.01 M

Moles of NaOH in 5.0 mL of 0.01 M NaOH = 0.01 × 5/1000 = 0.00005 moles

Moles of NaOH in 1.0 mL of 0.01 M NaOH = 0.01 ×1/1000 = 0.0001 moles

Ratio of moles of NaOH in 5.0 mL to 1.0 mL = 0.00005/0.00001 = 5

There are five times more moles of NaOH in 5.0 mL than in 1.0 mL and this means that 5 times more the quantity of HC₂H₃O2(aq) required to react with 1.0 mL NaoH is needed to react with 5.0 mL NaOH.

Therefore, the amount of HC₂H₃O2(aq) in the flask after the addition of 5.0mL of NaOH(aq) compared to the amount of HC₂H₃O₂(aq) in the flask after the addition of 1.0mL is much smaller because more HC₂H₃O₂(aq) is required to react with 5.0 mL NaOH than with 1.0 mL NaOH.

4 0

3 years ago

What is the structural formula of pent-3-en-1-yl? <br>Thx

krek1111 [17]

Happy birthday girl i’m love you too bye mommy love you mommy bye bye merry christmas

3 0

3 years ago

How many grams of solid Na2CO3 are required to neutralize exactly 2 liters of an HCI solution of pH 2.0?

Snezhnost [94]

Answer:

The answer is 1.06g.

Explanation:

Analysis of question:

1. Identify the information in the question given.

volume of HCl is 2 dm3
pH of HCl is 2.0

2. What the question want?

mass of Na2CO3 is ?(unknown)
3. Do calculation.
1st-Write a balanced chemical equation:

Na2CO3 + 2HCl (arrow) 2NaCl + H20 + CO2

2nd-Determine the molarity of HCl with the value of 2.0.

pH= -log[H+]

2.0= -log[H+]

log[H+]= -2.0

[H+]= 10 to the power of negative 2(10-2)

=0.01 mol dm-3

molarity of HCl is 0.01 mol dm-3

3rd-Find the number of moles of HCl

n=MV

=0.01 mol dm-3 × 2 dm3

=0.02 mol of HCl

4th-Find the second mol of it.

Based on the chemical equation,

2.0 mol of HCl reacts with 1.0 mol of Na2CO3

0.02 mol of HCl reacts with 0.01 mol of Na2CO3

<u>N</u><u>a</u>2CO3>a=<u>1</u><u> </u>mol

<u>2</u><u>H</u>Cl>b=<u>2</u><u> </u>mol

5th-Find the mass of it.

mass= number of mole × molar mass

g=0.01 × [2(23)+ 12+ 3(16)]

g=0.01 × 106

# =1.06 g.

3 0

3 years ago

At 298 K, what is the Gibbs free energy change (ΔG) for the following reaction?

9966 [12]

Answer:

(a). The Gibbs free energy change is 2.895 kJ and its positive.

(b). The Gibbs free energy change is 34.59 J/mole

(c). The pressure is 14924 atm.

(d). The Gibbs free energy of diamond relative to graphite is 4912 J.

Explanation:

Given that,

Temperature = 298 K

Suppose, density of graphite is 2.25 g/cm³ and density of diamond is 3.51 g/cm³.

$\Delta H\ for\ diamond = 1.897 kJ/mol$

$\Delta H\ for\ graphite = 0 kJ/mol$

$\Delta S\ for\ diamond = 2.38 J/(K mol)$

$\Delta S\ for\ graphite = 5.73 J/(K mol)$

(a) We need to calculate the value of $\Delta G$ for diamond

Using formula of Gibbs free energy change

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

Put the value into the formula

$\Delta G= (1897-0)-298\times(2.38-5.73)$

$\Delta G=2895.3$

$\Delta G=2.895\ kJ$

The Gibbs free energy change is positive.

(b). When it is compressed isothermally from 1 atm to 1000 atm

We need to calculate the change of Gibbs free energy of diamond

Using formula of gibbs free energy

$\Delta S=V\times\Delta P$

$\Delta S=\dfrac{m}{\rho}\times\Delta P$

Put the value into the formula

$\Delta S=\dfrac{12\times10^{-6}}{3.51}\times999\times10130$

$\Delta S=34.59\ J/mole$

(c). Assuming that graphite and diamond are incompressible

We need to calculate the pressure

Using formula of Gibbs free energy

$\beta= \Delta G_{g}+\Delta G+\Delta G_{d}$

$\beta=V(-\Delta P_{g})+\Delta G+V\Delta P_{d}$

$\beta=\Delta P(V_{d}-V_{g})+\Delta G$

Put the value into the formula

$0=\Delta P(\dfrac{12\times10^{-6}}{3.51}-\dfrac{12\times10^{-6}}{2.25})\times10130+2895.3$

$0=-0.0194\Delta P+2895.3$

$\Delta P=\dfrac{2895.3}{0.0194}$

$\Delta P=14924\ atm$

(d). Here, $C_{p}=0$

So, The value of $\Delta H$ and $\Delta S$ at 900 k will be equal at 298 K

We need to calculate the Gibbs free energy of diamond relative to graphite

Using formula of Gibbs free energy

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

Put the value into the formula

$\Delta G=(1897-0)-900\times(2.38-5.73)$

$\Delta G=4912\ J$

Hence, (a). The Gibbs free energy change is 2.895 kJ and its positive.

(b). The Gibbs free energy change is 34.59 J/mole

(c). The pressure is 14924 atm.

(d). The Gibbs free energy of diamond relative to graphite is 4912 J.

7 0

3 years ago

KClO3 --->KCl+3/2O2 assign oxidation states to each element on each side of the equation.(reactants and produc? KClO3 --->

marusya05 [52]

K will always have an oxidation state of +1. Now O is -2 except in peroxides, this is not a peroxide, so total charge will be -6, if you subtract the +1 of K from it, it leaves -5 charge to be neutralized by Cl in KClO3, so Cl will be +5. In the product side, K will still have the same oxdiation which is +1 and Cl would have -1. O2 will have zero. <span>Now, Cl is gaining the electrons to go from +5 to -1, so it is getting reduced while O2 is losing electrons to go from -2 to zero so it is getting oxidized.</span>

6 0

3 years ago