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zhannawk [14.2K]
3 years ago
14

What is the composition of the mantle

Chemistry
2 answers:
Olin [163]3 years ago
7 0
The mantle under the crust<span> is about 1,800 miles deep (2,890 km). It is composed mostly of </span>silicate rocks<span> rich in</span>magnesium<span> and iron. Intense heat causes the rocks to rise. They then cool and sink back down to the core.</span>
Sever21 [200]3 years ago
7 0

Answer:

The mantle is the middle layer of the earth, and it has a thickness of about 2900 km. It covers approximately 85% of the total earth's volume. This layer is comprised of elements such as oxygen, silicon, magnesium, aluminum, sodium, calcium, and iron. The rocks present here are mostly silicate rocks and are dominant in olivine, pyroxene, iron, and magnesium. These are denser minerals.

The mantle is again sub-divided into 2 distinct parts namely, upper mantle (includes the layer of asthenosphere) and lower mantle.

The upper mantle is comparatively less solid in nature in comparison to the lower mantle, which is marked by the presence of the asthenosphere layer, in which the rocks tend to flow in response to deformation.

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Is the prime meridian a longitudinal or latitudinal line?
Nesterboy [21]

Answer:

longitudinal

Explanation:

A prime meridian is the meridian (a line of longitude) in a geographic coordinate system at which longitude is defined to be 0°. Together, a prime meridian and its anti-meridian (the 180th meridian in a 360°-system) form a great circle. This great circle divides a spheroid into two hemispheres.

4 0
2 years ago
At the boiling point, the vapor pressure of a liquid is equal to the _______.
Marat540 [252]
At the boiling point, the vapor pressure of a liquid is equal to the atmospheric pressure.
6 0
3 years ago
Calculate the mass of CO2 that can be produced if the reaction of 54.0 g of propane and sufficient oxygen has a 64.0% yield.
Oduvanchick [21]

Answer:

103.9 g

Explanation:

  • C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

First <u>we convert 54.0 g of propane (C₃H₈) into moles</u>, using its <em>molar mass</em>:

  • 54.0 g ÷ 44 g/mol = 1.23 mol C₃H₈

Then we <u>convert 1.23 moles of C₃H₈ into moles of CO₂</u>, using the <em>stoichiometric coefficients</em>:

  • 1.23 mol C₃H₈ * \frac{3molCO_2}{1molC_3H_8} = 3.69 mol CO₂

We <u>convert 3.69 moles of CO₂ into grams</u>, using its <em>molar mass</em>:

  • 3.69 mol CO₂ * 44 g/mol = 162.36 g

And <u>apply the given yield</u>:

  • 162.36 g * 64.0/100 = 103.9 g
7 0
2 years ago
How many grams of lead(II) sulfate (303 g/mol) are needed to react with sodium chromate (162 g/mol) in order to produce 0.162 kg
Afina-wow [57]

Answer : The mass of PbSO_4 needed are, 1.515 grams.

Explanation :

First we have to calculate the mole of PbCrO_4.

\text{Moles of }PbCrO_4=\frac{\text{Mass of }PbCrO_4}{\text{Molar mass of }PbCrO_4}=\frac{0.162g}{323g/mole}=0.005mole

Now we have to calculate the moles of PbSO_4.

The balanced chemical reaction will be,

PbSO_4+Na_2CrO_4\rightarrow PbCrO_4+Na_2SO_4[tex]From the balanced chemical reaction, we conclude thatAs, 1 mole of [tex]PbCrO_4 produced from 1 mole of PbSO_4

So, 0.005 mole of PbCrO_4 produced from 0.005 mole of PbSO_4

Now we have to calculate the mass of PbSO_4

\text{Mass of }PbSO_4=\text{Moles of }PbSO_4\times \text{Molar mass of }PbSO_4

\text{Mass of }PbSO_4=0.005mole\times 303g/mole=1.515g

Therefore, the mass of PbSO_4 needed are, 1.515 grams.

6 0
2 years ago
Silver occurs in trace amounts in some ores of lead, and lead can displace silver from solution: Pb(s) + 2Ag+ (aq) LaTeX: \longr
VikaD [51]

Answer : The value of \Delta G^o and K is, -180 kJ/mol and 3.6\times 10^{31}

Explanation :

The balanced cell reaction will be,

Pb(s)+2Ag^+(aq)\rightarrow Pb^{2+}(aq)+2Ag(g)

The half-cell reactions are:

Oxidation reaction (anode) : Pb(s)\rightarrow Pb^{2+}(aq)+2e^-

Reduction reaction (cathode) : 2Ag^+(aq)+2e^-\rightarrow 2Ag(g)

Relationship between standard Gibbs free energy and standard electrode potential follows:

\Delta G^o=-nFE^o_{cell}

where,

\Delta G^o = standard Gibbs free energy

F = Faraday constant = 96500 C

n = number of electrons in oxidation-reduction reaction = 2

E^o_{cell} = standard electrode potential of the cell = 0.93 V

Now put all the given values in the above formula, we get:

\Delta G^o=-2\times 96500\times 0.93

\Delta G^o=-179490J/mol=-179.49kJ/mol\approx -180kJ/mol

Now we have to calculate the value of 'K'.

\Delta G^o=-RT\ln K

where,

\Delta G_^o =  standard Gibbs free energy  = -180 kJ/mol

R = gas constant = 8.314\times 10^{-3}kJ/mole.K

T = temperature = 298 K

K = equilibrium constant = ?

Now put all the given values in the above formula 1, we get:

-180kJ/mol=-(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln K

K=3.6\times 10^{31}

Therefore, the value of \Delta G^o and K is, -180 kJ/mol and 3.6\times 10^{31}

5 0
3 years ago
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