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marusya05 [52]
2 years ago
11

An ice cube tray full of water is put into a freezer. Which energy change occurs in the particles in the water as it undergoes a

phase change from a liquid to a solid?The potential energy decreases due to the tighter arrangement of the particles.
A.The potential energy decreases due to the tighter arrangement of the particles.

B.The potential energy increases due to the looser arrangement of the particles.

C..The potential energy decreases due to the looser arrangement of the particles.

D.The potential energy increases due to the tighter arrangement of the particles.
Chemistry
1 answer:
o-na [289]2 years ago
3 0

Answer:

A

Explanation:

increase in energy - hotter and/or more reactive

decrease in energy - colder and/or less reactive

Since it is going from a liquid to solid it will have a tighter arrangement of particles.

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a nuclide of 64/29 cu absorbs a positron. witch is the resulting atom? (A) 65/29Cu (B) 63/29 CU (C) 64/28Ni (D) 64/30 Zn
Nesterboy [21]

Answer : Option D) Zn^{64}_{30}

Explanation : When a positron is getting absorbed it means it will be e^{+1}_{0} so, the Cu^{64}_{29} will get converted;

So, the whole reaction will be;

Cu^{64}_{29} + e^{+1}_{0} ----> Zn^{64}_{30}.

This will convert the whole element of Cu will get changed into Zn. As, it absorbs by the positron, the atomic number gets increased from 29 to 30.

4 0
3 years ago
A 19.0 g piece of metal is heated to 99.0 °C and then placed in 150 mL of water at 21°C. The temperature of the water rose to 23
Anna35 [415]

Answer:

0.8696\ \text{J/g}^{\circ}\text{C}

Explanation:

m_m = Mass of metal = 19 g

c_m = Specific heat of the metal

\Delta T_m = Temperature difference of the metal = 99-23=76^{\circ}\text{C}

V = Volume of water = 150 mL = 150\ \text{cm}^3

\rho = Density of water = 1\ \text{g/cm}^3

c_w = Specific heat of the water = 4.186 J/g°C

\Delta T_w = Temperature difference of the water = 23-21=2^{\circ}\text{C}

Mass of water

m_w= \rho V\\\Rightarrow m_w=1\times 150\\\Rightarrow m_w=150\ \text{g}

Heat lost will be equal to the heat gained so we get

m_mc_m\Delta T_m=m_wc_w\Delta T_w\\\Rightarrow c_m=\dfrac{m_wc_w\Delta T_w}{m_m\Delta T_m}\\\Rightarrow c_m=\dfrac{150\times 4.186\times 2}{19\times 76}\\\Rightarrow c_m=0.8696\ \text{J/g}^{\circ}\text{C}

The specific heat of the metal is 0.8696\ \text{J/g}^{\circ}\text{C}.

4 0
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