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3 years ago

What type of weather does a warm front cause and why

1 answer:

7 0

Warm fronts often bring stormy weather as the warm air mass at the surface rises above the cool air mass, making clouds and storms.

Hope this helps!
Brainliest is much appreciated!

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Describe the development of the modern model of the atom.

umka21 [38]

Answer:

currently model of the atom shows an atom that is mostly empty space. In the center is a small nucleus made of protons and neutrons. The nucleus contains nearly all the mass of an atom. Surrounding the nucleus is a cloud-like region with electrons moving too fast and too unpredictably for us to know their location

Explanation:

3 0

3 years ago

Read 2 more answers

Determine the mass of 5.20 moles of C6H12 (gram-formula mass = 84.2 grams/mole).

SIZIF [17.4K]

5.20 mol C6H12* (84.2 g/mol C6H12)= 4.38*10^(2) g C6H12.

The answer should only have three significant figures, according to the numbers in the problem.

Hope this is helpful~

7 0

3 years ago

TRUE or FALSE: Most Tsunamis occur in the Atlantic ocean

ololo11 [35]

Answer:

true

Explanation:

4 0

4 years ago

A 80.0 g piece of metal at 88.0°C is placed in 125 g of water at 20.0°C contained in a calorimeter. The metal and water come to

grandymaker [24]

Answer:

The specific heat of the metal is 0.485 J/g°C

Explanation:

<u>Step 1:</u> Data given

Mass of the piece of metal = 80.0 grams

Mass of the water = 125 grams

Initial temperature of the metal = 88.0 °C

Initial temperature of water =20.0 °C

Final temperature = 24.7 °C

pecific heat of water is 4.18 J/g*°C

<u>Step 2:</u> Calculate specific heat of the metal

Qgained = -Qlost

Q =m*c*ΔT

Qwater = - Qmetal

m(water)*c(water)*ΔT(water) = -m(metal)*c(metal)*ΔT(metal)

with mass of water = 125 grams

with c( water) = 4.18 J/g°C

with ΔT(water) = T2-T1 = 24.7 - 20 = 4.7°C

with mass of metal = 80.0 grams

with c(metal) = TO BE DETERMINED

with ΔT(metal) = 24.7 - 88.0 = -63.3 °C

125*4.18*4.7 = -80 * C(metal) * -63.3

2455.75 = -80 * C(metal) * -63.3

C(metal) = 2455.75 / (-80*-63.3)

C(metal) = 0.485 J/g°C

The specific heat of the metal is 0.485 J/g°C

3 0

4 years ago

Please help me with this question, and if u can do it on paper

nasty-shy [4]

1. O2 is the limiting reagent

2. Excess NH3 = 0.102 moles

3. 0.088 mol of NO is formed

4. 2.376 grams of H2O is formed

<h3>Stoichiometric problem</h3>

From the equation: $4NH_3 + 5O_2 --- > 4NO + 6H_2O$

Mole ratio of NH3 and O2 = 4:5

Mole of 3.25 g NH3 = 3.25/17 = 0.19 mol

Mole of 3.50 g O2 = 3.5/32 = 0.11 mol

Equivalent mole of O2 from NH3 = 5/4 x 0.19 = 0.24 mol

Equivalent mole of NH3 from O2 = 4/5 x 0.11 = 0.088 mol

Thus, O2 is the limiting reagent.

Excess NH3 = 0.19 - 0.088 = 0.102 moles

Mole ratio of O2 and NO = 5:4

Equivalent mole of NO = 4/5 x 0.11 = 0.088 mol

Mole ratio of O2 and H2O = 5:6

Equivalent mole of H2O = 6/5 x 0.11 = 0.132 mol

Mass of 0.132 mol H2O = 0.132 x 18 = 2.376 grams

More on stoichiometric problems can be found here: brainly.com/question/15047541

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8 0

2 years ago