What is an energy source produced from nuclear reactions

Magma can partially crystallize at depth and then rise to shallow depths where the remaining magma solidifies. The early-formed

Doss [256]

<h2>Phenocrysts and Porphyritic texture </h2>

Explanation:

The early formed crystals are of phenocrysts and the texture of these crystals is porphyritic texture.
This crystallization occurs when early-forming plagioclase crystals which are rich in calcium start coating with plagioclase crystals which are rich in sodium.
On cooling, the magma is then processed in a volcanic eruption, after the eruption the liquid which is left behind will start cooling and forms a porphyritic texture.

5 0

3 years ago

When zinc or aluminum was allowed to react with the copper sulfate , what was the limiting reagen?

musickatia [10]

Usually it is the CuSO4 that is the limiting reagent.

if all of the color of the solution was gone, but there was still some zinc metal mixed in with the copper metal produced, then Zn is the excess reagent 

f all of the color of the solution was not gone, but there was no zinc metal left in with the blue copper solution , then Zn is the limiting reagent Hope this helps.

5 0

3 years ago

What kind of graph would use for this data? The months of the year and the amount of rainfall, and why

Digiron [165]

Answer:

A Desert Climatogram is a graph that shows the temperature and rainfall, for each month in the year, for the Desert.

Explanation:

6 0

3 years ago

PLZ HELP

shusha [124]

Use the equation for density :
Density = mass / volume
Density = 120 / 480
Density = 0.25

5 0

3 years ago

Read 2 more answers

Three 1.0-l flasks, maintained at 308 k, are connected to each other with stopcocks. initially the stopcocks are closed. one of

Licemer1 [7]

Answer:

0.6103 atm.

Explanation:

We need to calculate the vapor pressure of each component after the stopcocks are opened.

Volume after the stopcocks are opened = 3.0 L.

1) For N₂:

P₁V₁ = P₂V₂

P₁ = 1.5 atm & V₁ = 1.0 L & V₂ = 3.0 L.

P₂ of N₂ = P₁V₁ / V₂ = (1.5 atm) (1.0 L) / (3.0 L) = 0.5 atm.

2) For H₂O:

Pressure of water at 308 K is 42.0 mmHg.

we need to convert from mmHg to atm: (1.0 atm = 760.0 mmHg).

P of H₂O = (1.0 atm x 42.0 mmHg) / (760.0 mmHg) = 0.0553 atm.

We must check if more 2.2 g of water is evaporated,

n = PV/RT = (0.0553 atm) (3.0 L) / (0.082 L.atm/mol.K) (308 K) = 0.00656 mole.

m = n x cmolar mass = (0.00656 mole) (18.0 g/mole) = 0.118 g.

It is lower than the mass of water in the flask (2.2 g).

3) For C₂H₅OH:

Pressure of C₂H₅OH at 308 K is 102.0 mmHg.

we need to convert from mmHg to atm: (1.0 atm = 760.0 mmHg).

P of C₂H₅OH = (1.0 atm x 102.0 mmHg) / (760.0 mmHg) = 0.13421 atm.

We must check if more 0.3 g of C₂H₅OH is evaporated,

n = PV/RT = (0.13421 atm) (3.0 L) / (0.082 L.atm/mol.K) (308 K) = 0.01594 mole.

m = n x molar mass = (0.01594 mole) (46.07 g/mole) = 0.7344 g.

It is more than the amount in the flask (0.3 g), so the pressure should be less than 0.13421 atm.

We have n = mass / molar mass = (0.30 g) / (46.07 g/mole) = 0.00651 mole.

So, P of C₂H₅OH = nRT / V = (0.00651 mole) (0.082 L.atm/mole.K) (308.0 K) / (3.0 L) = 0.055 atm.

So, total pressure = P of N₂ + P of H₂O + P of C₂H₅OH = 0.5 atm + 0.0553 atm + 0.055 atm = 0.6103 atm.

3 0

3 years ago