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Furkat [3]
3 years ago
14

How much more powerful is a magnitude 8 earthquake than a magnitude 4 earthquake? A. 4 times B.100 times C.1000 times D.10000 ti

mes
Chemistry
1 answer:
Ede4ka [16]3 years ago
7 0

Answer:

C) 1,000

Explanation:

The answer is 1,000

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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
Andre45 [30]

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

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7 0
2 years ago
HELP ASAP!! PLEASE !!
ZanzabumX [31]

Answer:

8,3, 7,7

Explanation:

6 0
2 years ago
A solution is prepared by mixing 2.17 g of an unknown non-electrolyte with 225.0 g of chloroform. The freezing point of the resu
Deffense [45]

Answer:

The molar mass of the unknown non-electrolyte is 64.3 g/mol

Explanation:

Step 1: Data given

Mass of an unknown non-electrolyte = 2.17 grams

Mass of chloroform = 225.0 grams

The freezing point of the resulting solution is –64.2 °C

The freezing point of pure chloroform is – 63.5°C

kf = 4.68°C/m

Step 2: Calculate molality

ΔT = i*kf*m

⇒ ΔT = The freezing point depression = T (pure solvent) − T(solution) = -63.5°C + 64.2 °C = 0.7 °C

⇒i = the van't Hoff factor = non-electrolyte = 1

⇒ kf = the freezing point depression constant = 4.68 °C/m

⇒ m = molality = moles unknown non-electrolyte / mass chloroform

0.7 °C = 1 * 4.68 °C/m * m

m = 0.150 molal

Step 3: Calculate moles unknown non-electrolyte

molality = moles unknown non-electrolyte / mass chloroform

Moles unknown non-electrolyte = 0.150 molal * 0.225 kg

Moles unknown non-electrolyte = 0.03375 moles

Step 4: Calculate molecular mass unknown non-electrolyte

Molar mass = mass / moles

Molar mass = 2.17 grams / 0.03375 moles

Molar mass = 64.3 g/mol

The molar mass of the unknown non-electrolyte is 64.3 g/mol

6 0
3 years ago
7.20g of potassium sulfate reacts with the excess oxygen gas in a single displacement reaction, how many grams of oxygen are con
skad [1K]
The balanced chemical reaction:

K2SO4 + O2 = 2KO2 + SO2

Assuming that the reaction is complete, all of the potassium sulfate is consumed. We relate the substances using the chemical reaction. We calculate as follows:

7.20 g K2SO4 ( 1 mol / 174.26 g) ( 1 mol O2 / 1 mol K2SO4 ) ( 32 g / 1 mol ) = 1.32 g O2 consumed in the reaction.
5 0
3 years ago
The density of copper is listed as 8.94g/cm​ 3​ . Two students each make three density determinations of samples of the substanc
amm1812

Answer:

Density of the copper = 8.94g/cm^3

Student A results = 7.3gm/cm^3 ,9.4 gm/cm^3 , 8.3gm/cm^3

Student B results = 8.4 gm/cm^3 , 8.8 gm/cm^3 , 8gm/cm^3

From the observations we conclude that

Student A's result is accurate but not  precise as the trials noted are not close to each other.

Student B's result is accurate and precise as the trials noted are close to each other.

Mean density of student A = 7.3 + 9.4 + 8.3 /3 = 8.33gm/cm^3

Mean density of student B = 8.4 + 8.8 + 8 /3 = 8.4 gm/cm^3

both the densities of A and B are 0.5 away from the actual density.

5 0
3 years ago
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