Atmospheric pressure arises due to the force exerted by the air above the Earth. At higher altitudes, the mass of the air above the Earth is _____ than at sea level, and atmospheric pressure therefore _____ with altitude.


Answer 1


less, decreases


When the pressure of an atmosphere occurs because of the force exerted so at the time of the higher altitudes, the air mass i.e. above the earth should be less as the air is attracted towards surface of an earth because of the gravity and air contains the mass that shows near the surface area so automatically the air density reduced due to which the mass also decreased

Related Questions

Cyclohexanecarboxylic acid, C6H11COOH (pKa 4.90), is only slightly soluble in water, but its sodium salt, C6H11COO-Na , is quite soluble in water. Describe the solubility of cyclohexanecarboxylic acid in solutions of sodium hydroxide, sodium bicarbonate (NaHCO3), and sodium carbonate (Na2CO3). The pKa values for the conjugate acids of sodium hydroxide, sodium bicarbonate (NaHCO3), and sodium carbonate (Na2CO3) are 15.7, 6.36, and 10.33, respectively.
Unknown A melts at 113- 114oC. Known compounds 3-Nitroaniline and 4-Nitrophenol both melt at 112-114 oC. If A is mixed with 3-Nitroaniline and the melting point becomes broad and depressed, what must A be __________A) 3-Nitroaniline B) 4-Nitrophenol C) Both
This section of the periodic table is called a(n)
Consider the titration of 30 mL of 0.030 M NH3 with 0.025 M HCl. Calculate the pH after the following volumes of titrant have been added: a) 0 mL; b) 10 mL; c) 20 mL; d)35 mL; e) 36 mL; f) 37 mL.
Oil is a fossil fuel used to run cars, heat up homes, and produce electricity. Oil can be removed from the bottom of the ocean through drilling. Drilling machines dig deep down into the Earth. When oil is found, pipes carry it to the surface. Sometimes, accidents happen that cause the oil to spill into the oceans. The oil can not only kill marine organisms, but when it reaches the surface of the water, some chemicals evaporate, become part of the atmosphere, and pollute the air.Based on the passage, which resources and organisms are affected by oil spills? Select three options.a) airb) landc) waterd) birds in the aire) trees on landf) organisms in the water

As we drive an automobile, we dont't think about the chemical consumed and produced. Prepare a list of the principal chemicals consumed and produced during the operation of an automobile.




In an internal combustion engine operational in automobiles, fuels are converted into mechanical energy in order to move pistons.

The basic reaction in automobile engines is combustion.

     Principal chemicals consumed                 Chemicals produced

                   Petroleum                                           Carbon dioxide

                                                                               Carbon monoxide

                                                                               Nitrogen oxides

                                                                               Sulfur oxides

  In diesel engines, production of particulate carbon is also produced.

Automobiles such as car, truck, motorbikes run on petrol or diesel. While operating these automobiles, combustion of diesel or petrol takes place which in turn requires oxygen for the process to occur.

Operation of the automobiles consumes oxygen, petrol or diesel and releases harmful chemicals like carbon dioxide (CO2), sulphur dioxide (SO2), carbon monoxide (CO) and many more. These chemicals pollutes the air and also affects the survival of living organisms by affecting the respiratory organs.  

What is the mass of silver chlorate (191.32 g/mol) that decomposes to release 0.466L of oxygen gas at STP? AgC1036) _AgCl). _026) A) 1.33g B) 597 E) 7.968 C) 3.988 D) 2658


Answer : The mass of silver chlorate will be 2.654 grams.

Explanation :

The balanced chemical reaction is,

2AgClO_3\rightarrow 2AgCl+3O_2

First we have to calculate the moles of oxygen gas at STP.

As, 22.4 L volume of oxygen gas present in 1 mole of oxygen gas

So, 0.466 L volume of oxygen gas present in (0.466)/(22.4)=0.0208 mole of oxygen gas

Now we have to calculate the moles of silver chlorate.

From the balanced chemical reaction, we conclude that

As, 3 moles of oxygen produced from 2 moles of silver chlorate

So, 0.0208 moles of oxygen produced from (2)/(3)* 0.0208=0.01387 moles of silver chlorate

Now we have to calculate the mass of silver chlorate.

\text{Mass of }AgClO_3=\text{Moles of }AgClO_3* \text{Molar mass of }AgClO_3

Molar mass of silver chlorate = 191.32 g/mole

\text{Mass of }AgClO_3=0.01387mole* 191.32g/mole=2.654g

Therefore, the mass of silver chlorate will be 2.654 grams.

Enter the oxidation number of one atom of each element in each reactant and product.CH4(g)+2O2(g)⟶CO2(g)+2H2O(g)
C in CH4 :
H in CH4 :
O in O2 :
C in CO2 :
O in CO2 :
H in H2O :
O in H2O :
Which atom is reduced?
Which atom is oxidized?


The oxidation numbers of the atoms of the specified elements in each of the given atoms are;

1) -4

1) -42) +1

1) -42) +13) 0

1) -42) +13) 04) +4

1) -42) +13) 04) +45) -2

1) -42) +13) 04) +45) -26) +1

1) -42) +13) 04) +45) -26) +17) -2

1) -42) +13) 04) +45) -26) +17) -2Atom oxidized = C

1) -42) +13) 04) +45) -26) +17) -2Atom oxidized = CAtom reduced = O

1) C in CH4

To get the oxidation number of C;

Oxidation state of hydrogen atom is +1 and so if the oxidation state of C is x, then we have;

x + 4(+1) = 0

x + 4 = 0

x = -4

2) H in CH4

Oxidation state on Carbon atom in this case is -4. Thus;

-4 + 4x = 0

4x = 4

x = +1

3) O in O2

This is oxygen gas that exists in it's free state and as such oxidation number is 0.

4) C in CO2

Oxidation state of O here is -2. Thus;

x + 2(-2) = 0

x - 4 = 0

x = +4

5) O in CO2

Oxidation state of C is +4 here. Thus;

4 + 2x = 0

2x = -4

x = -4/2

x = -2

6) H in H2O

Oxidation state of oxygen here is -2. Thus;

2x - 2 = 0

2x = 2

x = 2/2

x = +1

7) O in H2O

Oxidation state of hydrogen here is +1. Thus;

2(1) + x = 0

x = -2

Finally, oxidation number of carbon increased, then it is the atom that was oxidized while the atom reduced is the Oxygen atom.

Read more at;


1. -4

2. +1

3. 0

4. +4

5. -2

6. +1

7. -2

reduced = H

oxidized = O


Know oxidation rules.

- Hope this helped! Please let me know if you would like to learn this. I could show you the rules and help you work through them.

Estimate how much heat in joules is released when 25.0 g of water (C = 4.184 J/g°C) is cooled from 80.0°C to 30.0°C?


The amount of heat will be 5230 j.

What is heat?

Heat is a type of energy that is transferred between both the system and its surroundings as a result of temperature variations.

Calculation of heat.

Given data:

Mass = 25.0 g = 0.025 kg

C = 4.184 J/g°C

T_(1) =  80.0°C

T_(2) = 30.0°C

Q= ?

By using the formula of heat.

Q = MC (T_(2) - T_(1))

Put the value of given data in heat equation.

Q(heat)  = 0.025  × 4.184  ( 30 - 80)

Q(heat) = 5230 J.

Therefore, the amount of heat will be 5230 J.

To know more about heat.



5230 J


m = 25 g = 0,025 kg

c = 4,184 J /(g * °C) = 4184 J /(kg * °C)

t_(1) = 80 °C

t_(2) = 30 °C

The formula is Q = c *m * (t_(2) - t_(1))


Q = 4184 * 0,025 * (30 - 80) = 5230 J

Note that we get a negative heat (-5230 J). It just means that it is released.

When an ion‑selective electrode for X+ was immersed in 0.0482 M XCl, the measured potential was 0.0460 V . What is the concentration of X+ when the potential is 0.0610 V ? Assume that the electrode follows the Nernst equation, the temperature is at 25 °C, and that the activity coefficient of X+ is 1.


Final answer:

To find the concentration of X+ at a potential of 0.0610V, use the Nernst equation which describes the electrochemical potential of a system. Given the initial concentration of X+ and its potential, rearrange the equation to solve for the concentration of X+ at the new potential.


The problem given can be solved using the Nernst equation, which relates the reduction potential of an electrochemicalreaction (half-cell or full cell reaction) to the standard electrode potential, temperature, and the activities of the chemical species undergoing the reduction.

The Nernst equation at 25 °C can be simplified as:

E = E° - (0.059/n) log [Cl^- /[X^+]

Where E is the electrode potential, E° is the standard electrode potential, n is the number of electron transferred and [Cl^- /[X^+] is the ratio of ion activities. Since the ion's activity coefficient is 1, we can treat [X^+] as the concentration of X^+.

If you apply this equation, using the given potentials and known initial concentration of X^+, you can solve for the concentration of X^+ when the potential is 0.0610V.

Learn more about Nernst Equation here:


Final answer:

The concentration of X+ can be calculated using the Nernst equation by substituting the initial and final potentials. By solving the equation, you can find the concentration of X+.


The concentration of X+ can be calculated using the Nernst equation. The Nernst equation relates the potential of a cell to the concentration of the ions involved.

The Nernst equation is given by:

E = Eº - (0.0592/n)log(Q)

Where E is the potential, Eº is the standard electrode potential, n is the number of electrons transferred, and Q is the reaction quotient.

In this case, the initial potential is 0.0460 V and the final potential is 0.0610 V. By substituting these values into the Nernst equation, you can solve for the concentration of X+.

Learn more about Nernst equation here:


A buffer solution is 0.413 M in HF and 0.237 M in KF. If Ka for HF is 7.2×10-4, what is the pH of this buffer solution?





Any buffer system can be described with the reaction:


Where HA is the acid and A^- is the base. Additionally, the calculation of the pH of any buffer system can be made with the Henderson-Hasselbach equation:


With all this in mind, we can write the reaction for our buffer system:


In this case, the acid is HF with a concentration of 0.413 M and the base is F^- with a concentration of 0.237 M. We can calculate the pKa value if we do the "-Log Ka", so:


Now, we can plug the values into the Henderson-Hasselbach


The pH value would be 2.90

I hope it helps!