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Naily [24]
2 years ago
12

The gravitational force exerted by an object is given by F = mg, where F is the force in newtons, m is the mass in kilograms, an

d g is the acceleration due to gravity (9.81 m/s²).
(b) Osmium (Z = 76) is a transition metal in Group 8B(8) and has the highest density of any element (22.6 g/mL). If an osmium column is 1 m² in area, how high must it be for its pressure to equal atmospheric pressure? [Use the answer from part (a) in your calculation.]
Chemistry
1 answer:
Andre45 [30]2 years ago
7 0

The height of the column is 0.457 m and the mass of the atmosphere is calculated as 1.03 × 10 ⁴ kg.

<h3>What is Pascal? </h3>

Pascal is defined as the force per unit area. It expressed in Newton pr square meter of area.

1 Pa = 1 N / m²

Pressure = force / Area

According to the question, the expression of force is as given below

F = mg

where,

F is the force,

m is the mass,

g is equal to the acceleration due to the gravity

Now, the area of the atmosphere is 1 m².The pressure is 1 atm. Pressure in Pascal.

1 atm = 1.01325 × 10 5pa

Therefore, the expression for pascal become as follows.

1.01325 × 10 5 pa = mg /area

1.01325 × 10 5 pa = m × 9.81 m/s² / 1 m²

M = 1.01325 × 10 5 pa × 1 m² / 9.81 m² × 1 Nm -² /1 pa × 1 kg m-² / 1 N

1.03 × 10 ⁴ kg

Given,

The density is 22.6 g /mL , pressure is 1 atm, and area is 1 m²

The relation between density and pressure can be given as follows.

P = hpg… … …(1 )

were , h is the height of the column

p is the density.

Hpg = 1.01325 × 10 5 pa × 1 N/m² /1 pa

H = 1.01325 × 10 5 N/m² / pg × 1 kg ms-² / 1 N

= 1.01325 × 10 5 kg m-¹ s -² / 22.6 g mL -1 × 1 kg/ 10 ³ g × 1 mL / 10 -6 m ³ × 9.81 m s- ²

= 0.457 m

Therefore, the height of the column is 0.457 m.

Thus, we concluded that the height of the column is 0.457 m and the mass of the atmosphere is calculated as 1.03 × 10 ⁴ kg.

learn more about density:

brainly.com/question/952755

#SPJ4

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Answer:

T_{eq}=28.9\°C

Explanation:

Hello!

In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:

Q_{Cd}+Q_{W}=0

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m_{Cd}C_{Cd}(T_{eq}-T_{Cd})+m_{W}C_{W}(T_{eq}-T_{W})=0

In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:

T_{eq}=\frac{m_{Cd}C_{Cd}T_{Cd}+m_{W}C_{W}T_{W}}{m_{Cd}C_{Cd}+m_{W}C_{W}}

Now, we plug in to obtain:

T_{eq}=\frac{37.60g*0.232\frac{J}{g\°C}*100.00\°C+25.00g*4.184\frac{J}{g\°C}*23.0\°C}{37.60g*0.232\frac{J}{g\°C}+25.00g*4.184\frac{J}{g\°C}}\\\\T_{eq}=28.9\°C

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