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xenn [34]
3 years ago
15

List three criteria used to identify a cold front.

Chemistry
1 answer:
zaharov [31]3 years ago
3 0

Answer:

Explanation:

A cold front is a transition zone from warm air masses to cold air masses. We use weather elements to identify cold fronts.

  1. Temperature: Since air moves from warm masses to cold masses, the air in a cold front is at different temperatures. The air behind a cold front is warm while the one ahead is cold. This implies that within a cold front is at different temperatures.
  2. Precipitation: Prior to the passing of cold front, precipitation is usually rife with low showers. A coldfront in itself is usually accompanied by heavy rainfall full of thunderstorms and lightening. After a coldfront, the showers steadies and decreases.
  3. Pressure changes: before a coldfront, the atmospheric pressure decreases steadily. When the front arrives, the pressure further lowers with a sharp increase thereafter. After the front, the pressure can continue to increase.
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A beaker of water has a volume of 125mL and a density of 1.0g/mL. Calculate the mass of the water.​
liq [111]

Answer:     <em><u>125 g</u></em>

Explanation:

Mass = Volume × Density

⇒ Mass of the water = 125 mL × 1.0 g/mL

                                  =  125 g

7 0
3 years ago
Why does tire pressure change with temperature? Explain why the atoms are interacting in the certain way, How temperature plays
Wittaler [7]

The main reason behind this is Boyle's law and Gay Lussacs law

  • Gay Lussac's law states that Pressure is directly proportional to Temperature.
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If temperature will be more Pressure will be more

6 0
3 years ago
A 1.00-L sample of carbon tetrachloride has a mass of 1.58 kg. What is the density of this substance in g/cm3?
Hoochie [10]

Answer:

1.58gcm⁻³

Explanation:

Given parameters:

Volume of sample = 1.00L

Mass of sample = 1.58kg

unknown:

Density of the sample

Solution:

The density of a substance is expressed as its mass per unit volume.

The units of the parameters are given for volume as liters and for mass as kg.

We need to convert to cm³ for volume and into g for the mass:

        1L = 1000cm³:

        Therefore, the volume of the sample is 1000cm³

For the mass:

           1kg = 1000g

           1.58kg gives 1.58 x 1000g = 1580g

The density of the sample will be:

               Density  = \frac{mass}{volume}

               Density = \frac{1580}{1000} = 1.58gcm⁻³

3 0
3 years ago
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7 0
3 years ago
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A certain water tower hold 4.6×10^5 gallons of water. find the volume of that water in liters.​
Readme [11.4K]

Answer: 1741289L

Explanation:

1 gallon = 3.78541 L

4.6×10^5 gallons = 4.6×10^5 x 3.78541 = 1741289L

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