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27 days ago

In the same condition, what happens to the net force acting on the rope?

Chemistry

1 answer:

lorasvet [2.5K]27 days ago

3 0

Response:

Outlined below

Clarification:

According to Newton's first law of motion, an object at rest will stay at rest, and one in motion will maintain that state unless acted upon by a net external force.

Hence, when two equal and opposite forces are applied to the rope, it experiences zero net force and remains stationary. Therefore, the rope is balanced and remains at rest; if it was already moving, it would continue in motion without interruption.

On the other hand, if the forces apply in the same direction, it creates an unbalanced force, leading to a net force equal to the total of the two. This results in a stronger force than the individual components, causing a stationary rope to move or stopping a moving rope.

Greetings!

The result is:

The new volume is: $2L$

Rationale:

Because the temperature remains constant, we can apply Boyle's Law to solve this issue.

Boyle's Law stipulates that:

$P_{1}V_{1}=P_{2}V_{2}$

Where,

P is the gas's pressure.

V is the gas's volume.

According to the information provided:

$V_{1}=2.5L\\P_{1}=6.0atm\\P_{2}=7.5atm$

Let's put the values into the equation:

$2.5L*6.0atm=7.5atm*V_{2}$

$2.5L*6.0atm=7.5atm*V_{2}\\\\V_{2}=\frac{2.5L*6.0atm}{7.5atm}=\frac{15L.atm}{7.5atm}=2L$

Consequently, the new volume is: $2L$

Wishing you a lovely day!

7 0

1 month ago

PART A: Use the following glycolytic reaction to answer the question. If the concentration of DHAP is 0.125 M and the concentrat

alisha [2704]

Answer:

For A: The change in free energy for the reaction is -5339.76 J/mol

For B: Free energy change is expressed in kJ/mol

For C: The forward reaction favors progression, while the reverse reaction does not.

Explanation:

Regarding the specified chemical reaction:

$DHAP\rightleftharpoons G_3P$

For A:

The relationship between standard Gibbs free energy and equilibrium constant is as follows:

$\Delta G^o=-RT\ln K_{eq}$

The free energy change can be calculated using the following equation:

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

Or,

$\Delta G=-RT^o\ln K_{eq}+RT\ln Q$

where,

$\Delta G$ = Change in free energy

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

$T^o$ = standard temperature = $25^oC=[273+25]K=298K$

T = temperature of the cell = $37^oC=[273+37]K=310K$

$K_[eq}$ = equilibrium constant = $5.4\times 10^{-2}$

Q = reaction quotient = $\frac{[G_3P]}{[DHAP]}$

$[G_3P]$ = 0.06 M

[DHAP] = 0.125 M

Substituting the values into the equation yields:

$\Delta G=[-(8.314J/mol.K\times 298K\times \ln (5.4\times 10^{-2}))]+[(8.314J/mol.K\times 310K\times \ln (\frac{0.06}{0.125}))]\\\\\Delta G=-[-7231.46]+[-1891.7]=-5339.76J/mol$

Thus, the change in free energy for the reaction is -5339.76 J/mol

For B:

To convert the free energy change to kilojoules, we apply the conversion factor:

1 kJ = 1000 J

So, $-5339.76J/mol\times \frac{1kJ}{1000J}=-5.34kJ/mol$

Consequently, the free energy change's units are kJ/mol

For C:

For spontaneity in the reaction, the Gibbs free energy must be negative. However, the calculations indicate a positive Gibbs free energy, leading to the conclusion that the reaction is not spontaneous.

The free energy change of the reaction is negative.

Consequently, the forward reaction is favored and the reverse reaction is not favored.

8 0

27 days ago

Transamination of an amino acid transfers an amine group to form an α‑keto acid and is catalyzed by transaminases. Some amino ac

lorasvet [2515]

The amino acids classified under the first group include alanine, aspartate, and glutamate, whereas those in the second group consist of glycine, valine, proline, leucine, isoleucine, methionine, serine, threonine, cysteine, asparagine, glutamine, phenylalanine, tryptophan, tyrosine, lysine, arginine, and histidine.

7 0

20 days ago

The normal boiling point of c2cl3f3 is 47.6°c and its molar enthalpy of vaporization is 27.49 kj/mol. what is the change in entr

eduard [2509]

Based on the equation:

ΔG = ΔH - TΔS = 0

It follows that ΔS = ΔH/T

So, ΔS = n*ΔHVap / Tvap

- where n represents the number of moles calculated as mass/molar mass

For a mass of 24.1 g

and a molar mass of 187.3764 g/mol

substituting gives:

∴ n = 24.1 / 187.3764g/mol

= 0.129 moles

The molar enthalpy of vaporization, ΔHvap, is 27.49 kJ/mol

The temperature in Kelvin, Tvap = 47.6 + 273 = 320.6 K

After substitution, we compute ΔS, the change in entropy:

∴ΔS = 0.129 mol * 27490 J/mol / 320.6 K

= 11 J/K

7 0

29 days ago

1. Which two groups of elements in the periodic table are the most reactive?(1 point) A) alkali metals and halogens B) alkaline

alisha [2704]

Answer:

Explanation:

1) Alkali metals and halogens both need to achieve a stable outer electron shell, requiring alkali metals to lose one electron and halogens to gain one.

2) They share an identical count of outer shell electrons.

3) Typically, they have elevated melting points.

4) They exhibit low reactivity or none at all.

5) They belong to group 7.

4 0

1 month ago