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torisob [31]
3 years ago
6

Which is an example of kinetic energy?

Chemistry
2 answers:
masya89 [10]3 years ago
4 0

Answer:

B

Explanation:

I'm not sure.,jsjajjwkwkpasbxbaakjwiw

professor190 [17]3 years ago
4 0

A

I m not sure ...........................

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Explain why a balloon filled with O2 at room temperature collapses when cooled in liquid nitrogen where as balloon filled with H
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When cooled by liquid nitrogen, the balloon shrinks (not as much as the air-filled balloon) and it sinks down on the table. When heating up, the balloon slowly rises and flies up in the air again. Explanation 1: The volume of the balloon decreases by the low temperature, because the gas inside is cooled down.
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How much heat energy is needed to raise the temperature of 59.7g of cadmium from 25°C to 100°C? The specific heat of cadmium is
labwork [276]
     Using the Fundamental Equation of Calorimetry, we have:

Q=mc\Delta T \\ Q=59.7.0.231.(100-25) \\ \boxed {Q=1034.3025J}      

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Metal cations are atoms that have a positive charge. What type of bond is a metal cation likely to form? A. covalent bond B. dou
Lunna [17]

Answer:

D. ionic bond

Explanation:

Due to electron deficiency in a metal cation, they cannot form a covalent bond beacuse it means to share electrons. By contrast metal cation seek for electrons. In an ionic bond, one atom give electrons, while another atom recevie electron. Because of that, this is the better option to metal cations.

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5. How much heat (in calories) is absorbed by a reaction when
Wewaii [24]

Answer:

1.2 × 10⁴ cal

Explanation:

Given data

  • Mass of water (m): 300 g
  • Initial temperature: 80 °C
  • Final temperature: 40°C

We can calculate the heat released by the water (Q_w) when it cools using the following expression.

Q_w = c \times m \times (T_f - T_i)

where

c is the specific heat capacity of water (1 cal/g.°C)

Q_w = \frac{1cal}{g.\°C}  \times 300g \times (40\°C - 80\°C) = -1.2 \times 10^{4} cal

According to the law of conservation of energy, the sum of the heat released by the water (Q_w) and the heat absorbed by the reaction (Q_r) is zero.

Q_w + Q_r = 0\\Q_r = -Q_w = 1.2 \times 10^{4} cal

7 0
3 years ago
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