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valentina_108 [34]
3 years ago
14

When one thermal equilibrium achieved between two objects ?

Chemistry
2 answers:
Elodia [21]3 years ago
6 0
Hello! Thermal equilibrium is reached between two objects when both have the same final heat value or temperature. Meaning when ice is dropped into water, and the ice melts, the ice and water mixture will get to thermal equilbrum when the temp of the water has stabilized and the ice added has completely melted and warmed up to the water temp so ice and water are now in thermal equilibrium.
Hope this helps you understand the concept. Any questions please just ask. Thank you!!
Setler [38]3 years ago
4 0
Based on the Zeroth Law of Thermodynamics, two systems are at equilibrium when there is no heat flow, q, between the systems.
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1 One mole of an element contains which of the following?
lozanna [386]

Answer:

1) B    2) both a and c

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3 years ago
How many moles of sodium hydroxide are there in 1.0mL of 2.0M NaOH
kvasek [131]
To find the moles, you can use the following formula

moles= Molarity x Liters

Molarity= 2.0 M
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4 0
2 years ago
An excited ozone molecule, O3*, in the atmosphere can undergo one of the following reactions,O3* → O3 (1) fluorescenceO3* → O +
Maurinko [17]

Answer:

The simplified expression for the fraction  is  \text {X} =    \dfrac{  {k_3  \times cM} }{k_1 +k_2 + k_3 }

Explanation:

From the given information:

O3* → O3                   (1)    fluorescence

O + O2                      (2)    decomposition

O3* + M → O3 + M    (3)     deactivation

The rate of fluorescence = rate of constant (k₁) × Concentration of reactant (cO)

The rate of decomposition is = k₂ × cO

The rate of deactivation = k₃ × cO × cM

where cM is the concentration of the inert molecule

The fraction (X) of ozone molecules undergoing deactivation in terms of the rate constants can be expressed by using the formula:

\text {X} =    \dfrac{ \text {rate of deactivation} }{ \text {(rate of fluorescence) +(rate of decomposition) + (rate of deactivation) }  } }

\text {X} =    \dfrac{  {k_3 \times cO \times cM} }{  {(k_1 \times cO) +(k_2 \times cO) + (k_3 \times cO \times cM) }  }

\text {X} =    \dfrac{  {k_3 \times cO \times cM} }{cO (k_1 +k_2 + k_3  \times cM) }

\text {X} =    \dfrac{  {k_3  \times cM} }{k_1 +k_2 + k_3  }    since  cM is the concentration of the inert molecule

7 0
3 years ago
On a periodic table, element names are represented by:
alina1380 [7]

Answer: Symbols

Explanation:

4 0
3 years ago
1. Ethanol and 2,6-dimethylphenol both have hydroxyl groups that can be deprotonated by a base. Why does ethanol act as a solven
Volgvan

Anions are solvated in protic hydrogen-bonding solvents (such as ethanol). Consequently, nucleophiles are less reactive. Since soft nucleophiles are less strongly solvated than hard nucleophiles, these solvents boost the relative reactivity of soft anions.

<h3>Ethanol is either a nucleophile or a base.</h3>

The ethanol is a base  Because carbocation is an extremely reactive species, a base or nucleophile as weak as ethanol can replace or remove it. SN1 and E1 would not be conceivable without the carbocation or a strong departing group.

<h3>How do solvents impact anionic nucleophile's reactivity?</h3>

In polar aprotic solvents, nucleophilic substitution reactions of anionic nucleophiles often proceed more quickly. The normal relative reactivity order in such solvents (like DMSO)is Anions are solvated in protic hydrogen-bonding solvents (such as ethanol). Consequently, nucleophiles are less reactive.

Learn more about nucleophiles here:-

brainly.com/question/27127109

#SPJ4

3 0
1 year ago
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