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2 years ago

How does energy from Earth’s interior affect surface changes?

2 answers:

6 0

Earth's rotation influences global flow of air and water. 2.3 Earth's weather and climate are mostly driven by energy from the Sun. For example, unequal warming of Earth's surface and atmosphere by the Sun drives convection within the atmosphere, producing winds, and influencing ocean currents.

quester [9]2 years ago

5 0

Answer:

Energy produced deep inside Earth heats rock in the mantle. ... As it becomes less dense, the heated rock rises toward Earth's surface. The cooler, denser rock surrounding the heated rock sinks, as Figure 5 shows. In this way, heat inside Earth moves toward the cooler crust.

Explanation:

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The chemical equation, Cr + Fe(NO3)2 → Fe + Cr(NO3)3, is an example of which type of reaction?

Angelina_Jolie [31]

Answer:

Redox type

Explanation:

The reaction is:

2Cr + 3Fe(NO₃)₂ → 2Fe + 2Cr(NO₃)₃

2 moles of chromium can react to 3 moles of iron (II) nitrate in order to produce 2 moles of iron and 2 moles of chromium nitrate.

If we see oxidation state, we see that chromium changes from 0 to +3

Iron changed the oxidation state from +2 to 0

Remember that elements at ground state has 0, as oxidation state.

Iron is being reduced while chromium is oxidized. Then, the half reactions are:

Fe²⁺ + 2e⁻ ⇄ Fe (Reduction)

Cr ⇄ Cr³⁺ + 3e⁻ (Oxidation)

When an element is being reduced, while another is being oxidized, we are in prescence of a redox reaction.

8 0

3 years ago

Explain three factors that can affect the size of a line in a spectrum.

Sunny_sXe [5.5K]

Answer:For atoms and molecules, the width of spectral lines is governed mainly by the broadening of the energy levels of the atoms or molecules during interactions with surrounding particles and by the broadening of the spectral lines as a result of the Doppler effect.

Explanation:

7 0

2 years ago

Onsider the following reaction at equilibrium:

11111nata11111 [884]

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

For the given equation:

$C(s)+H_2O(l)\leftrightharpoons CO(g)+H_2(g)$

A. C is added to the reaction mixture.

If the concentration of reactant specie is increased , according to the Le-Chatlier's principle , the equilibrium will shift in the direction where decrease in concentration of reactant specie occurs. So, on increasing C ,equilibrium will shift in right direction.

B. $H_2O$ is condensed and removed from the reaction mixture.

If the concentration of reactant specie is decreased , according to the Le-Chatlier's principle , the equilibrium will shift in the direction where increase in concentration of reactant specie occurs. So, on removing water vapor ,the equilibrium will shift in left direction.

C. CO is added to the reaction mixture.

If the concentration of product specie is increased , according to the Le-Chatlier's principle , the equilibrium will shift in the direction where decrease in concentration of product specie occurs. So, on increasing CO, the equilibrium will shift in left direction.

D. $H_2$ is removed from the reaction mixture.

If the concentration of product specie is decreased , according to the Le-Chatlier's principle , the equilibrium will shift in the direction where increase in concentration of product specie occurs. So, on removing hydrogen , the equilibrium will shift in right direction.

5 0

3 years ago

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l), ΔH = –1.37 × 103 kJ For the combustion of ethyl alcohol as described in the above equati

babymother [125]

Answer:

The true statements are: I. The reaction is exothermic.

II. The enthalpy change would be different if gaseous water was produced.

Explanation:

The given chemical reaction: C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l), ΔH= -1.37×10³kJ

1. In an exothermic reaction, heat or energy is released from the system to the surrounding. Thus for an exothermic process the change in enthalpy is less than 0 or negative (ΔH < 0) .

Since the enthalpy change for a combustion reaction is negative. Therefore, the given reaction is exothermic.

2. The change in enthalpy (ΔH) of a reaction is equal to difference of the sum of standard enthalpy of formation (ΔHf°) of the products and the reactants.

ΔHr° = ∑ n.ΔHf°(products) − ∑ n.ΔHf°(reactants)

As the value of ΔHf° of water in gaseous state and liquid state is not the same.

Therefore, the enthalpy change of the reaction will be different, if gaseous water was present instead of liquid water.

3. An oxidation-reduction reaction or a redox reaction involves simultaneous reduction and oxidation processes.

The given chemical reaction, represents the combustion reaction of ethanol.

Since combustion reactions are redox reactions. Therefore, the given combustion reaction is an oxidation-reduction reaction.

4. According to the ideal gas equation: P.V =n.R.T

Volume (V) ∝ n (number of moles of gas)

Since the number of moles (n) of gaseous reactants is 3 and number of moles of gaseous (n) products is 2.

Therefore, the volume occupied by 3 moles of the reactant gaseous molecules will be more than 2 moles product gaseous molecules.

3 0

3 years ago

(I)how many atoms are present in 7g of lithium?

ICE Princess25 [194]

Answer :

(i) The number of atoms present in 7 g of lithium are, $6.07\times 10^{23}$

(ii) The number of atoms present in 7 g of lithium are, $1.204\times 10^{24}$

(iii) The number of moles of $F_2$ is, 1 mole

The number of moles of $CO_2$ is, 0.5 mole

The number of moles of $OH^-$ is, 1 mole

Explanation :

Part (i) :

First we have to calculate the moles of lithium.

$\text{Moles of }Li=\frac{\text{Mass of }Li}{\text{Molar mass of }Li}$

Molar mass of Li = 6.94 g/mole

$\text{Moles of }Li=\frac{7g}{6.94g/mol}=1.008mole$

Now we have to calculate the number of atoms present.

As, 1 mole of lithium contains $6.022\times 10^{23}$ number of atoms

So, 1.008 mole of lithium contains $1.008\times 6.022\times 10^{23}=6.07\times 10^{23}$ number of atoms

Thus, the number of atoms present in 7 g of lithium are, $6.07\times 10^{23}$

Part (ii) :

First we have to calculate the moles of carbon.

$\text{Moles of }C=\frac{\text{Mass of }C}{\text{Molar mass of }C}$

Molar mass of C = 12 g/mole

$\text{Moles of }C=\frac{24g}{12g/mol}=2mole$

Now we have to calculate the number of atoms present.

As, 1 mole of carbon contains $6.022\times 10^{23}$ number of atoms

So, 2 mole of carbon contains $2\times 6.022\times 10^{23}=1.204\times 10^{24}$ number of atoms

Thus, the number of atoms present in 7 g of lithium are, $1.204\times 10^{24}$

Part (iii) :

To calculate the moles of $F_2$ :

$\text{Moles of }F_2=\frac{\text{Mass of }F_2}{\text{Molar mass of }F_2}$

Molar mass of $F_2$ = 38 g/mole

$\text{Moles of }F_2=\frac{19g}{19g/mol}=1mole$

Thus, the number of moles of $F_2$ is, 1 mole

To calculate the moles of $CO_2$ :

$\text{Moles of }CO_2=\frac{\text{Mass of }CO_2}{\text{Molar mass of }CO_2}$

Molar mass of $CO_2$ = 44 g/mole

$\text{Moles of }CO_2=\frac{22g}{44g/mol}=0.5mole$

Thus, the number of moles of $CO_2$ is, 0.5 mole

To calculate the moles of $OH^-$ ions :

$\text{Moles of }OH^-=\frac{\text{Mass of }OH^-}{\text{Molar mass of }OH^-}$

Molar mass of $OH^-$ = 17 g/mole

$\text{Moles of }OH^-=\frac{17g}{17g/mol}=1mole$

Thus, the number of moles of $OH^-$ is, 1 mole

4 0

3 years ago