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boyakko [2]
3 years ago
15

Based on the description of geothermal energy, how does magma convection aid in the production of geothermal energy?

Chemistry
2 answers:
Ratling [72]3 years ago
6 0

Answer:

Magma convection moves heat from Earth’s deep interior up to the crust, where it can be used to produce energy.

Explanation:

Delvig [45]3 years ago
5 0

Explanation:

Magma convection is the primary driving mechanism by which geothermal energy is brought to the earth.

Geothermal energy is energy from deep within earth has been brought close to crustal levels.

They are usually sourced from areas in which a body of magma is appreciably close to the surface or where the mantle is close up.

  • Mantle convection is movement of magma within the region of the earth's mantle.
  • In this process, hot magma rises up close to the surface and cold ones sinks further deep into the mantle.
  • As hot magma rises in the mantle, it is brought close to the surface and crustal levels.
  • Where the crust is thin, the heat from the ensuing magma can be felt more close up
  • Cracks in the overlying lithosphere can bring these magma up ensuring a continuous supply of this hot mantle materials.

learn more:

Renewable resources brainly.com/question/2948717

#learnwithBrainly

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What is the percent yield for a process in which 10.4g of CH3OH reacts and 10.1 g of CO2 is formed
monitta

Answer:

A. 70.7%

Explanation:

In the first step lets compute the molar mass of CH₃OH and CO

Molar Mass of CH₃OH =  1(12.01 g/mol) + 4(1.008 g/mol) +1(16.00 g/mol)

                                     = 32.042 g/mol

Molar Mass of CO₂      = 1(12.01 g/mol) + 2(16.00 g/mol)  

                                     = 44.01 g/mol

                                   

Mass of only one reactant i.e. CH₃OH is given so  it must be the limiting reactant. Next, the theoretical yield is calculated directly as follows:

Given mass of CH₃OH is 10.4 g. So we have:

                                     10.4g CH₃OH

Convert grams of CH₃OH to moles of CH₃OH utilizing molar mass of CH₃OH as:

                          1 mol CH₃OH / 32.042 g CH₃OH

Convert CH₃OH to moles of CO₂ using mole ratio as:

                             2 mol CO₂ / 2 mol CH₃OH

Convert moles of  CO₂ to grams of  CO₂ utilizing molar mass of  CO₂ as:

                           44.01 g/mol CO₂ / 1 mol CO₂

Now calculating theoretical yield using above steps:

[ 10.4 g CH₃OH ]  [1 mol CH₃OH / 32.042 g CH₃OH ]  [2 mol CO₂ / 2 mol CH₃OH]  [44.01 g/mol CO₂ / 1 mol CO₂]

Multiplication is performed here. We are left with 10.4 and 44.01 g CO₂ from numerator terms in the above equation and 32.042 from denominator terms after cancellation process of above terms. So this equation becomes:

= ( 10.4 ) ( 44.01 ) g CO₂ / 32.042

= 457.704/32/042

=  14.28 g CO₂

Theoretical yield =  14.28 g CO₂  

Finally compute the percent yield for a process in which 10.4g of CH₃OH reacts and 10.1 g of CO₂ is formed:

percent yield = (actual yield / theoretical yield) x 100

As we have calculated theoretical yield which is 14.28 g CO₂ and actual yield is 10.1 g CO₂ So,

percent yield = (10.1 g CO₂ / 14.28 g CO₂) x 100%

                       = 0.707 x 100%

                       = 70.7 %

Hence option A 70.7% yield is the correct answer.

8 0
3 years ago
Most Bic lighters hold 5.0ml of liquified butane (density = 0.60 g/ml). Calculate the minimum size container you would need to "
Hatshy [7]

Answer:

Volume of container = 0.0012 m³ or 1.2 L or 1200 ml

Explanation:

Volume of butane = 5.0 ml

density = 0.60 g/ml

Room temperature (T) = 293.15 K

Normal pressure (P) = 1 atm = 101,325 pa

Ideal gas constant (R) = 8.3145 J/mole.K)

volume of container V = ?

Solution

To find out the volume of container we use ideal gas equation

PV = nRT

P = pressure

V = volume

n = number of moles

R = gas constant

T = temperature

First we find out number of moles

<em>As Mass = density × volume</em>

mass of butane = 0.60 g/ml ×5.0 ml

mass of butane = 3 g

now find out number of moles (n)

n = mass / molar mass

n = 3 g / 58.12 g/mol

n = 0.05 mol

Now put all values in ideal gas equation

<em>PV = nRt</em>

<em>V = nRT/P</em>

V = (0.05 mol × 8.3145 J/mol.K × 293.15 K) ÷ 101,325 pa

V = 121.87 ÷ 101,325 pa

V = 0.0012 m³ OR 1.2 L OR 1200 ml

8 0
3 years ago
Please help!!
Zinaida [17]
1.B
6.D 
D because t<span>he mass of one mole (molar mass) of helium gas is </span>4.002602 g/mol. 4.002602 * 5=20.01309. Rounded equals 20. So, the answer is D.20 g.
8 0
3 years ago
A sample of P 2 O 4 contains 181 . g of phosphorus. Calculate the mass of the sample in pounds
-Dominant- [34]

Answer:

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7 0
3 years ago
Give the electronic configuration of carbon atom.​
zimovet [89]

Answer:

1s2 2s2 2p2

Explanation:

it has 6 electrons in two energy levels so the sub levels are 1s, 2s and 2p

4 0
3 years ago
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