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4 years ago

Explain how heat in earths interrior is transferred

1 answer:

7 0

There are three process by which the heats are transferred.They are conduction,convection & radiation. The concept of thermos flax designed to avoid all these process,that is why we are able to keep at required temp without loss are gain of temp. The convection is a method by which the heat is transfer through third media.Please look in to my site for earth section.The heat at mandrill is transferred to lithosphere and to crest by the medium called magma. This is carrying the heat from inner layer to outer core of the earth.How the egg immersed in hot water is boiled is the same process.

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A sample of h gas (2.0 l) at 3.5 atm was combined with 1.5 l of n gas at 2.6 atm pressure at a constant temperature of 25c into

nadya68 [22]

For this problem, assume ideal gas then we use the Combined Gas Law.

P₁V₁/T₁ + P₂V₂/T₂ = PV/T
where P, V and T are the total pressure, volume and temperature

Since temperature is kept constant, the equation is simplified to:

P₁V₁ + P₂V₂ = PV
(3.5 atm)(2L) + (2.6 atm)(1.5 L) = (P)(7 L)
Solving for P,
P = 1.56 atm

3 0

3 years ago

What is the temperature in kelvin of the Xe in the balloon?

katrin [286]

A2.37 l balloon contains .115 mol of xenon gas, xe(g), at a pressure of 954 torr. what is the temperature, in kelvin, of the xe in the balloon? express the answer using 3 significant figures.

3 0

4 years ago

What patterns in the rock layer fossils helped you identify an extinction?

Lilit [14]

Answer:

Hope this helps!! :D

Explanation:

Often, the rock layers bookending the mass extinction are noticeably different in their compositions. These changes in the rocks show the effects of environmental disturbances that triggered the mass extinction and sometimes hint at the catastrophic cause of the extinction.

3 0

3 years ago

Read 2 more answers

Calculate the energy (in kj/mol) required to remove the electron in the ground state for each of the following one-electron spec

Bess [88]

Explanation:

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number

a) Energy change due to transition from n = 1 to n = ∞ ,hydrogen atom .

Z = 1

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{1^2}{(\infty)^2}eV=0$

Let energy change be E for 1 atom.

$E=E_{\infty}-E_1=0-(-13.6 eV)=13.6 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 13.6 eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 13.6 \times 1.60218\times 10^{-22} kJ/mol$

$E'=1,312.17 kJ/mol$

The energy required to remove the electron in the ground state is 1,312.17 kJ/mol.

b) Energy change due to transition from n = 1 to n = ∞ , $B^{4+}$ atom .

Z = 5

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{5^2}{1^2}eV=-340 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{5^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-340eV)=340 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 340eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 340\times 1.60218\times 10^{-22} kJ/mol$

$E'=32,804.31 kJ/mol$

The energy required to remove the electron in the ground state is 32,804.31 kJ/mol.

c) Energy change due to transition from n = 1 to n = ∞ , $Li^{2+}$ atom .

Z = 3

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{3^2}{1^2}eV=-122.4 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{3^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-122.4 eV)=122.4 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 122.4 eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 122.4\times 1.60218\times 10^{-22} kJ/mol$

$E'=11,809.55 kJ/mol$

The energy required to remove the electron in the ground state is 11,809.55 kJ/mol.

d) Energy change due to transition from n = 1 to n = ∞ , $Mn^{24+}$ atom .

Z = 25

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{25^2}{1^2}eV=-8,500 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{25^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-8,500 eV)=8,500 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 8,500eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 8,500 \times 1.60218\times 10^{-22} kJ/mol$

$E'=820,107.88 kJ/mol$

The energy required to remove the electron in the ground state is 820,107.88 kJ/mol.

4 0

4 years ago

What happens at the anode and the cathode of an electrochemical cell? Why do electrons flow from the anode to the cathode?

Lelechka [254]

Explanation:

An electrochemical cell is a device that converts chemical energy to electrical energy by generating potential difference through the eletrodes. It is composed of various units like the anode, cathode, salt bridge, voltmeter etc.

At the anode, the negative electrode at which the oxidation half-reaction occurs, that is the positive ions are reduced.

At the cathode, the positive electrode at which the reduction half-reaction occurs.

Example of an Ionic equation;

At the cathode (Reduction),

Y+(aq) + e- --> Y(aq)

At the anode (Oxidation),

Z-(aq) --> Z(aq) + e-

Electrons always flow from the anode (negative electrode) to the cathode (positive electrode) or from the oxidation half cell to the reduction half cell, as ssen in the illustrative example above.

In terms of E°cell of the half reactions, the electrons will flow from the more negative half reaction to the more positive half reaction.

3 0

4 years ago