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2 years ago

Which parameter is indicative of how far and in what direction a reaction will progress?

Chemistry

1 answer:

nekit [7.7K]2 years ago

6 0

Answer:

Option b, The change in free energy of the reaction (ΔG)

Explanation:

Gibbs free energy is a measure of amount of usable energy in the system.

It is related with enthalpy (H), entropy (S) and temprature (T) as:

G = H - TS

The Gibbs free energy change (ΔG) provide spontaneity of a chemical reaction.

If ΔG is negative, then reaction is spontaneous that means reaction is moving towards forward direction.

If ΔG is positive, then reaction is non-spontaneous that means reaction is moving in backward direction.

If ΔG is zero, then reaction is at equilibrium.

Change in enthalpy only gives informtion about heat involed in a chemical reaction, it does not give information about direction of the reaction.

So, among the given options, option b is correct.

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What happens when an atom has more electrons than can fit in one energy

Rina8888 [55]

Answer:

If any atom has more electrons than one energy level can hold, then automatically the electron is accommodated in the next energy level (shell). The remaining extra electrons starts to fill the next energy level. This produces the valency of that particular atom.

Explanation:

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3 years ago

The reaction of aluminum with bromine is shown here. The equation for the reaction is

m_a_m_a [10]

Answer is (b)

2,3,1 is the stoichiometry

The values in front of the elements are the stoichiometric values
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2 years ago

According to the law of conservation of matter, we know that the total number of atoms does not change in a chemical reaction an

Shkiper50 [21]

Answer:

Compounds are represented by chemical formulas.

Explanation:

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2 years ago

Are the properties of Rubidium more similar to those of cesium or those of strontium?

fenix001 [56]

More similar to Cesium

Explanation:

The properties of Rubidium are more similar to those of cesium compared to strontium.

Elements in the same group on the periodic table have similar chemical properties.

Rubidium and Cesium are located in the first group on the periodic table.
Other elements in this group are lithium, sodium, potassium and francium
Strontium belongs to the second group on the periodic table.
The first group have a ns¹ valence shells electronic configuration.
They are all referred to as alkali metals

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Sodium brainly.com/question/6324347

Periodic table brainly.com/question/1704778

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2 years ago

A 100 gram glass container contains 200 grams of water and 50.0 grams of ice all at 0°c. a 200 gram piece of lead at 100°c is ad

ASHA 777 [7]

$0 \; \textdegree{\text{C}}$

Explanation:

Assuming that the final (equilibrium) temperature of the system is above the melting point of ice, such that all ice in the container melts in this process thus

$E(\text{fusion}) = m(\text{ice}) \cdot L_{f}(\text{water}) = 66.74 \; \text{kJ}$ and
$m(\text{water, final}) = m(\text{water, initial}) + m(\text{ice, initial}) = 0.250 \; \text{kg}$

Let the final temperature of the system be $t \; \textdegree{\text{C}}$ . Thus $\Delta T (\text{water}) = \Delta T (\text{beaker}) = t(\text{initial}) - t_{0} = t \; \textdegree{\text{C}}$

$Q(\text{water}) &= &c(\text{water}) \cdot m(\text{water, final}) \cdot \Delta T (\text{water})= 1.047 \cdot t\; \text{kJ}$ (converted to kilojoules)
$Q(\text{container}) &= &c(\text{glass}) \cdot m(\text{container}) \cdot \Delta T (\text{container})= 0.0837 \cdot t \; \text{kJ}$
$Q(\text{lead}) &= &c(\text{lead}) \cdot m(\text{lead}) \cdot \Delta T (\text{lead})= 0.0255 \cdot (100 - t)\; \text{kJ}$

The fact that energy within this system (assuming proper insulation) conserves allows for the construction of an equation about variable $t$ .

$E(\text{absorbed} ) = E(\text{released})$

$E(\text{absorbed} ) = E(\text{fushion}) + Q(\text{water}) + Q(\text{container})$
$E(\text{released}) = Q(\text{lead})$

Confirm the uniformity of units, equate the two expressions and solve for $t$ :

$66.74 + 1.047 \cdot t + 0.0837 \cdot t = 0.0255 \cdot (80 - t)$

$t \approx -55.95\; \textdegree{\text{C}} < 0\; \textdegree{\text{C}}$ which goes against the initial assumption. Implying that the final temperature does <em>not</em> go above the melting point of water- i.e., $t \le 0 \; \textdegree{\text{C}}$ . However, there's no way for the temperature of the system to go below $0 \; \textdegree{\text{C}}$ ; doing so would require the removal of heat from the system which isn't possible under the given circumstance; the ice-water mixture experiences an addition of heat as the hot block of lead was added to the system.

The temperature of the system therefore remains at $0 \; \textdegree{\text{C}}$ ; the only macroscopic change in this process is expected to be observed as a slight variation in the ratio between the mass of liquid water and that of the ice in this system.

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2 years ago