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4 years ago

Rain puddle drying up, chemical or physical change

Chemistry

1 answer:

Volgvan4 years ago

3 0

Physical change because it's getting smaller as it disappears.

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How many electrons can be held in sub level I=3?

Zinaida [17]

2 is your answer hope you get it right

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3 years ago

How many moles are present when there are 1.00.1026 molecules of water?

liraira [26]

Answer: 1.00×10^26=1000×10^23

1000×10^23/6.022×10^23=166.05×10^23 moles of water.

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Explanation:

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3 years ago

A container of argon with a volume of 22 ml has a pressure of 101.0 kPa.

Alexeev081 [22]

Answer:

The new pressure is 44.4 kPa.

Explanation:

We have,

Initial volume, $V_1=22\ ml$

Initial pressure, $P_1= 101.0\ kPa$

It is required to find the new pressure when the volume is increased to 50 ml. The relationship between pressure and volume is known as Boyle's law.

$PV=k\\\\P_1V_1=P_2V_2$

$P_2$ is final pressure

$P_2=\dfrac{P_1V_1}{V_2}\\\\P_2=\dfrac{22\times 101\times 10^3}{50}\\\\P_2=44440\ Pa\\\\P_2=44.4\ kPa$

So, new pressure is 44.4 kPa.

6 0

3 years ago

A piece of unknown metal with mass 30 g is heated to 110.0 °C and dropped into 100.0 g of water at 20.0 °C. The final temperatur

Ymorist [56]

<u>Answer:</u> The specific heat of metal is 0.821 J/g°C

<u>Explanation:</u>

When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of metal = 30 g

$m_2$ = mass of water = 100 g

$T_{final}$ = final temperature = 25°C

$T_1$ = initial temperature of metal = 110°C

$T_2$ = initial temperature of water = 20.0°C

$c_1$ = specific heat of metal = ?

$c_2$ = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

$30\times c_1\times (25-110)=-[100\times 4.186\times (25-20)]$

$c_1=0.821J/g^oC$

Hence, the specific heat of metal is 0.821 J/g°C

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3 years ago

Why do planets in our solar system stay in their positions

Anon25 [30]

This is not chemistry but it's A, off top.

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4 years ago

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