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____ [38]
4 years ago
15

How many more times intense is an earthquake that measures 8 on the Richter scale than an earthquake that measures 5? Explain yo

ur answer.
Chemistry
2 answers:
Tema [17]4 years ago
6 0

Answer: The earthquake will be 1000 times more intense when it is measured 8 on the Richter scale.

Explanation: Richter scale is the numerical scale which measures the intensity of earthquake with the help of seismographic oscillations.

Each level on the Richter scale is 10 times stronger than the previous one.

Richter scale measuring 8 is three levels up than the value 5 on the scale.

Hence, the earthquake having 8 on the Richter scale will be ( 10 × 10 × 10) = 1000 times more intense than the earthquake having 5 on the Richter scale.

djyliett [7]4 years ago
3 0

The answer on edguinity is A magnitude 8 earthquake is 1,000 times more intense than a magnitude 5 earthquake. A magnitude 8 earthquake is 108 times more intense than a standard earthquake, while a magnitude 5 earthquake is 105 times more intense than a standard earthquake, and 108 ÷ 105 = 103. Each unit increase on the Richter scale corresponds to an intensity increase by a factor of 10. So from 5 to 8 on the Richter scale, the intensity increases by 103 = 1,000.

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What is the equation for SHC? Give all<br> units
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Q=mcΔT

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What is used for mixing a small amount of chemicals together?
DENIUS [597]

a thin solid glass rod that is used in chemistry to combine substances. A stirring rod often has rounded ends and is about the length of a long straw.

<h3>What use serves the stirring rod?</h3>

A crucial component of lab apparatus for mixing chemicals and liquids for reactions is a long, thin stirring rod. Stirring rods are made of solid plastic, glass, or steel and are non-abrasive, chemically inert, and chemically resistant.

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7 0
2 years ago
Determine the freezing point and boiling point of a solution that has 68.4 g of sucrose
Ymorist [56]

Answer:

Freezing T° of solution = - 3.72°C

Boiling T° of solution =  101.02°C

Explanation:

To solve this we apply colligative properties. Firstly, freezing point depression:

ΔT = Kf . m . i

ΔT = Freezing T° of pure solvent - Freezing T° of solution

Kf = Cryoscopic constant, for water is 1.86 °C/m

m = molality (moles of solute in 1kg of solvent)

i = Ions dissolved in solution

Our solute is sucrose, an organic compound so no ions are defined. i = 1.

Let's determine the moles: 68.4 g . 1mol/ 342g = 0.2 moles

molality = 0.2 mol / 0.1kg of water = 2 m

We replace data: ΔT = 1.86°C/m . 2m . 1

Freezing T° of solution = - 3.72°C

Now, we apply elevation of boiling point: ΔT = Kb . m . i

ΔT = Boiling T° of solution - Boiling T° of  pure solvent

Kf = Ebulloscopic constant, for water is 0.512 °C/m

We replace:

Boiling T° of solution - Boiling T° of pure solvent = 0.512 °C/m . 2 . 1

Boiling T° of solution = 0.512 °C/m . 2 . 1 + 100°C → 101.02°C

6 0
3 years ago
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