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Fittoniya [83]
2 years ago
5

At the _______ point of a substance, the particles have enough kinetic energy to break free from their fixed positions.

Chemistry
1 answer:
Natasha_Volkova [10]2 years ago
5 0

Answer:

At the nucleus point of a substance, the particles have enough kinetic energy to break free from their fixed positions.

Answer : nucleus

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When you look out the window you can see constantly moving air.Air is cycled in order to reach equilibrium
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False, although the air is moving in an equilibrium us normal humans can not see the air actually moving. So it is false.
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What are ionic compounds. Explain.<br>Explain what is gravity.​
Sveta_85 [38]

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6 0
3 years ago
Read 2 more answers
Para formar bronce, se mezclan 150g de cobre a 1100°C y 35g de estaño a 560°C. Determine la temperatura final del sistema.
jek_recluse [69]

Answer:

La temperatura final del sistema es 1029,346 °C.

Explanation:

Asumamos que el sistema conformado por el cobre y el estaño no tiene interacciones con sus alrededores. Por la Primera Ley de la Termodinámica, el cobre cede calor al estaño con tal de alcanzar el equilibrio térmico. El cobre se encuentra inicialmente en su punto de fusión, mientras que el estaño está por encima de ese punto, de modo que la transferencia de calor es esencialmente sensible:

m_{Cu}\cdot c_{Cu}\cdot (T-T_{Cu}) = m_{Sn}\cdot c_{Sn}\cdot (T_{Sn}-T)

(m_{Cu}\cdot c_{Cu} + m_{Sn}\cdot c_{Sn})\cdot T = m_{Sn}\cdot c_{Sn}\cdot T_{Sn} + m_{Cu}\cdot c_{Cu}\cdot T_{Cu}

T = \frac{m_{Sn}\cdot c_{Sn}\cdot T_{Sn}+m_{Cu}\cdot c_{Cu}\cdot T_{Cu}}{m_{Cu}\cdot c_{Cu}+m_{Sn}\cdot c_{Sn}} (1)

Donde:

m_{Sn} - Masa del estaño, en gramos.

m_{Cu} - Masa del cobre, en gramos.

c_{Sn} - Calor específico del estaño, en calorías por gramo-grados Celsius.

c_{Cu} - Calor específico del cobre, en calorías por gramo-grados Celsius.

T_{Sn} - Temperatura inicial del estaño, en grados Celsius.

T_{Cu} - Temperatura inicial del cobre, en grados Celsius.

Si sabemos que m_{Cu} = 150\,g, m_{Sn} = 35\,g, c_{Cu} = 0,093\,\frac{cal}{g\cdot ^{\circ}C}, c_{Sn} = 0,060\,\frac{cal}{g\cdot ^{\circ}C}, T_{Sn} = 560\,^{\circ}C y T_{Cu} = 1100\,^{\circ}C, entonces la temperatura final del sistema es:

T = \frac{(35\,g)\cdot \left(0,060\,\frac{cal}{g\cdot ^{\circ}C} \right)\cdot (560\,^{\circ}C)+(150\,g)\cdot \left(0,093\,\frac{cal}{g\cdot ^{\circ}C} \right)\cdot (1100\,^{\circ}C)}{(35\,g)\cdot \left(0,060\,\frac{cal}{g\cdot ^{\circ}C} \right)+(150\,g)\cdot \left(0,093\,\frac{cal}{g\cdot ^{\circ}C} \right)}

T = 1029,346\,^{\circ}C

La temperatura final del sistema es 1029,346 °C.

3 0
3 years ago
The following neutral atoms each have 8 places for electrons in their outermost shells.
Lunna [17]
C. ne with 8 valenc3 electrons
7 0
3 years ago
Read 2 more answers
At a certain concentration of H2 and NH3, the initial rate of reaction is 0.120 M / s. What would the initial rate of the reacti
mel-nik [20]

The question is incomplete, here is the complete question:

The rate of certain reaction is given by the following rate law:

rate=k[H_2]^2[NH_3]

At a certain concentration of H_2 and [tex]I_2, the initial rate of reaction is 0.120 M/s. What would the initial rate of the reaction be if the concentration of [tex]H_2 were halved.Answer : The initial rate of the reaction will be, 0.03 M/sExplanation :Rate law expression for the reaction:[tex]rate=k[H_2]^2[NH_3]

As we are given that:

Initial rate = 0.120 M/s

Expression for rate law for first observation:

0.120=k[H_2]^2[NH_3] ....(1)

Expression for rate law for second observation:

R=k(\frac{[H_2]}{2})^2[NH_3] ....(2)

Dividing 2 by 1, we get:

\frac{R}{0.120}=\frac{k(\frac{[H_2]}{2})^2[NH_3]}{k[H_2]^2[NH_3]}

\frac{R}{0.120}=\frac{1}{4}

R=0.03M/s

Therefore, the initial rate of the reaction will be, 0.03 M/s

5 0
3 years ago
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