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3 years ago

What type of front is associated with hail storms

Chemistry

2 answers:

Gekata [30.6K]3 years ago

7 0

I think its a cold front.

lana [24]3 years ago

5 0

many fronts cause weather events like rain, thunderstorms, gusty winds, and tornadoes. at a warm front there may be low stratus clouds. usually the skies clear after the front has passed,a weather front is a border between two different air masses at the earth's surface

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a gas has a volume of 350 cubic centimeters at 740 mmHg. how many cubic centimeters will the gas occupy at a pressure of 900 mmH

Harrizon [31]

Answer: 287.8 cm3

Explanation:

Given that:

Initial volume of gas V1 = 350 cm3

Initial pressure of gas P1 = 740 mmHg

New volume V2 = ?

New pressure P2 = 900 mmHg

Since, pressure and volume are involved while temperature is constant, apply the formula for Boyle's law

P1V1 = P2V2

740 mmHg x 350 cm3 = 900mmHg x V2

V2 = (740 mmHg x 350 cm3) /900mmHg

V2 = 259000 mmHg cm3 / 900mmHg

V2 = 287.8 cm3

Thus, the gas will occupy 287.8 cubic centimeters at the new pressure.

3 0

3 years ago

energy that comes from the heat of the earth's core is known as: solar energy hydro energy geothermal energy wave energy

Alecsey [184]

Energy that comes from the heat of the Earth's core is known as geothermal energy.
Solar energy comes from the Sun, hydro energy from water, and wave energy from wind and waves.

7 0

3 years ago

Read 2 more answers

How many moles (of molecules or formula units) are in each sample? 79.34 g cf2cl2?

Alona [7]

From the periodic table:
molecular mass of carbon = 12 grams
molecular mass of fluorine = 18.99 grams
molecular mass of chlorine = 35.5 grams
Therefore:
one mole of CF2Cl2 = 12 + 2(18.99) + 2(35.5) = 120.98 grams
Therefore, we can use cross multiplication to find the number of moles in 79.34 grams as follows:
mass = (79.34 x 1) / 120.98 = 0.6558 moles

Now, one mole contains 6.022 x 10^23 molecules, therefore:
number of molecules in 0.65548 moles = 0.6558 x 6.022 x 10^23
= 3.949 x 10^23 molecules

7 0

3 years ago

Help help help help help help

natita [175]

Hey,

Answer:

AgCl (Silver Chloride)

Explanation:

sodium chloride + silver nitrate → sodium nitrate+ silver chloride

Hope it helps :)

Have a great day!

3 0

2 years ago

A dark brown binary compound contains oxygen and a metal. It is 13.38% oxygen by mass. Heating it moderately drives off some of

Leto [7]

Answer:

a) Mass of O in compound A = 32.72 g

Mass of O in compound B = 21.26 g

Mass of O in compound C = 15.94 g

b) Compound A = MO2

Compound B = M3O4

Compound C = MO

c) M = Pb

Explanation:

Step 1: Data given

A binairy compound contains oxygen (O) and metal (M)

⇒ 13.38 % O

⇒ 100 - 13.38 = 86.62 % M

After heating we get another binairy compound

⇒ 9.334 % O

⇒ 100 - 9.334 = 90.666 % M

After heating we get another binairy compound

⇒ 7.168 % O

⇒ 100 - 7.168 = 92.832 % M

The first compound has an empirical formula of MO2

⇒ 1 mol M for 2 moles O

Step 2: Calculate amount of metal and oxygen in each

compound A: M = m1 *0.8662 O = m1 *0.1338

compound B: M = m2 *0.90666 O = m2 *0.09334

compound C: M = m3 *0.92832 O = m3 *0.07168

Step 3: Calculate mass of oxygen with 1.000 grams of M

Compound A: 1.000g * 0.1338 m1gO / 0.8662m1gMetal = 0.1545

Compound B: 1.000g * 0.09334 m2gO / 0.90666m2gMetal = 0.1029

Compound C: 1.000g * 0.07168 m3gO / 0.92832m3gMetal = 0.07721

Step 4:

1 mol MO2 has 1 mol M and 2 moles O

m1 = (mol O * 16)/0.1338 m1 = 239.2 grams

1 mol M = 0.8632*239.2 = 206.48

0.90666m2 = 206.48 ⇒ m2 = 227.74 g

0.92832m3 = 206.48 ⇒ m3 = 222.42 g

Step 5: Calculate mass of O

Mass of O in compound A = 239.2 - 206.48 = 32.72 g

Mass of O in compound B = 227.74 - 206.48 = 21.26 g

Mass of O in compound C = 222.42- 206.48 = 15.94 g

Step 6: Calculate moles

Moles of O in compound A ≈ 2

⇒ MO2

Moles of O in compound B = 21.26 / 16 ≈ 1.33

⇒ M3O4

Moles of O compound C = 15.94 /16 ≈ 1 moles

⇒ MO

Step 7: Calculate molar mass

The mass of 1 mol metal is 206.48 grams ⇒ molar mass ≈ 206.48 g/mol

The closest metal to this molar mass is lead (Pb)

6 0

3 years ago