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3 years ago

If the pressure of a 2.00 L sample of gas is 50.0 kPa, what pressure does the gas exert if its volume is decreased to 20.0 mL?

Chemistry

1 answer:

dimulka [17.4K]3 years ago

8 0

Answer:

P₂ = 5000 KPa

Explanation:

Given data:

Initial volume = 2.00 L

Initial pressure = 50.0 KPa

Final volume = 20.0 mL (20/1000=0.02 L)

Final pressure = ?

Solution:

The given problem will be solved through the Boly's law,

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume

Now we will put the values in formula,

P₁V₁ = P₂V₂

50.0 KPa × 2.00L = P₂ × 0.02 L

P₂ = 100 KPa. L/0.02 L

P₂ = 5000 KPa

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Answer:

Number of delocalized electrons

Explanation:

Magnesium has more delocalized electrons compared to sodium and this accounts for the higher melting point.

When magnesium atoms comes together to form a metallic bonds, they have more network of delocalized electrons.
There is more pull for the localized electrons due to the nuclear charge on the nucleus.
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3 years ago

A 37.2 g sample of copper at 99.8 °C is carefully placed into an insulated container containing 188 g of water at 18.5 °C. Calcu

klasskru [66]

Answer:

T₂ = 19.95°C

Explanation:

From the law of conservation of energy:

$Heat\ Lost\ by\ Copper = Heat\ Gained\ by\ Water\\m_cC_c\Delta T_c = m_wC_w\Delta T_w$

where,

mc = mass of copper = 37.2 g

Cc = specific heat of copper = 0.385 J/g.°C

mw = mass of water = 188 g

Cw = specific heat of water = 4.184 J/g.°C

ΔTc = Change in temperature of copper = 99.8°C - T₂

ΔTw = Change in temperature of water = T₂ - 18.5°C

T₂ = Final Temperature at Equilibrium = ?

Therefore,

$(37.2\ g)(0.385\ J/g.^oC)(99.8\ ^oC-T_2)=(188\ g)(4.184\ J/g.^oC)(T_2-18.5\ ^oC)\\99.8\ ^oC-T_2 = \frac{(188\ g)(4.184\ J/g.^oC)}{(37.2\ g)(0.385\ J/g.^oC)}(T_2-18.5\ ^oC)\\\\99.8\ ^oC-T_2 = (54.92) (T_2-18.5\ ^oC)\\54.92T_2+T_2 = 99.8\ ^oC + 1016.02\ ^oC\\\\T_2 = \frac{1115.82\ ^oC}{55.92}$

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3 years ago

Water bugs walk on water, some razor blades can float on water, and water forms droplets. Are these examples of viscosity, surfa

Anton [14]

C.) Surface Tension, Because water can hold a certain amount of pressure/weight before the object sinks

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3 years ago

List two processes from which it may be concluded that the particles of a gas move continuously.

ziro4ka [17]

Answer:

The two processes maybe:

One process can be the spread of smell of perfume . Tge particles of perfume mic in the air and reaches every corner of tge room because they are moving

Second can be the smell of food that reaches us in seconds . This case too tells that the aroma of the food particles gets intermixed and come to us .

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4 years ago

You calculate the Wilson equation parameters for the ethanol (1) 1 1-propanol (2) system at 258C and find they are L12 5 0.7 and

Gala2k [10]

Here is the correct question.

You calculate the Wilson equation parameters for the ethanol (1) +1 - propanol (2) system at 25° C and find they are ∧₁₂ = 0.7 and ∧₂₁ = 1.1 . Estimate the value of parameters at 50° C

Answer:

the values of the parameters at 50° C are 0.766 and 1.047

Explanation:

From "critical point , enthalpy of phase change and liquid molar volume " the liquid molar volume (v) of ethanol and 1 - propanol is represented as follows:

Compound Liquid molar volume (cm³/mol)

Ethanol (1) 58.68

1 - propanol (2) 75.14

To calculate the temperature dependent parameters of the Wilson equation ∧₁₂.

∧₁₂ = $\frac{V_2}{V_1} \ exp \ (\frac{-a_{12}/R}{T} )$ ------------ equation (1)