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.

Answers

Answer 1
Answer:

Answer:

less, decreases

Explanation:

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


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Use the following structures of amino acids to answer the questions below. Note that the difference in the structures (the side chains) is highlighted by gray shading.A student performed chromatography of the four amino acids and theresults were shown in the chromatogram below. If an anion exchangecolumn (column is positively charged) was used in a neutral buffer,assign each amino acid to the corresponding peak in the chromatogram.

This section of the periodic table is called a(n)

Answers

Answer:

Is it Group?

Explanation:

Group 2A (or IIA) of the periodic table are the alkaline earth metals: beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba), and radium (Ra).

Match the words in the left column to the appropriate blanks in the sentences on the right. Make certain each sentence is complete before submitting your answer. ResetHelp 1. NaCl ; ionic bonds{\rm NaCl} ; blank are stronger than the blank in {\rm HCl}. are stronger than the dispersion forces{\rm NaCl} ; blank are stronger than the blank in {\rm HCl} . in HCl. 2. H2O ; hydrogen bonds{\rm H_2O} ; blank are stronger than the blank in {\rm H_2Se}. are stronger than the dispersion forces{\rm H_2O} ; blank are stronger than the blank in {\rm H_2Se}. in H2Se. 3. NH3 ; hydrogen bonds{\rm NH_3} ; blank are stronger than the blank in {\rm PH_3}. are stronger than the dipole-dipole attractions{\rm NH_3} ; blank are stronger than the blank in {\rm PH_3}. in PH3. 4. HF ; hydrogen bonds{\rm HF} ; blank are stronger than the blank in {\rm F_2}. are stronger than the dispersion forces{\rm HF} ; blank are stronger than the blank in {\rm F_2}. in F2.

Answers

Answer:

The ionic bond in NaCl are stronger than the stronger than the dispersion forces in HCl.

The hydrogen bonds in H2O are stronger than the dispersion forces in H2Se

Hydrogen bonds in NH3 are stronger than the dipole-dipole attractions in PH3.

Hydrogen bonds in HF are stronger than the dispersion forces in F2

Explanation:

Ionic bonds occur in molecules with high differences in their electronegative value where there are actual transfer of electrons. HCl has a bond which is involved in the sharing of electrons.

Hydrogen bonds are present in H2O which is stronger than the dispersion forces.

PH3 is a larger molecule with greater dispersion forces than ammonia, NH3 has very polar N-H bonds leading to strong hydrogen bonding. This dominant intermolecular force results in a greater attraction between NH3 molecules than there is between PH3 molecules.

F2 is a non-polar molecule, therefore they have London dispersion forces between molecules while HF has a hydrogen bond because F is highly electronegative.

Final answer:

Ionic bonds are stronger than dispersion forces in HCl, while hydrogen bonds are stronger than dispersion forces in H2Se, PH3, and F2.

Explanation:

In the given sentences, the blanks represent the types of intermolecular forces. The options given are ionic bonds, hydrogen bonds, dispersion forces, and dipole-dipole attractions. Ionic bonds are stronger than the dispersion forces in HCl. Hydrogen bonds are stronger than the dispersion forces in H2Se. Hydrogen bonds are stronger than the dipole-dipole attractions in PH3. Hydrogen bonds are stronger than the dispersion forces in F2.

Learn more about Intermolecular forces here:

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indigestion tablets neutralise acid in the stomach. what does this tell you about indigestion tablets?

Answers

Answer:

It is basic.

Explanation:

Bases can neutralize acids.

A Se ion has a mass number of 79 and a charge of − 2 . Determine the number of neutrons, protons, and electrons in this ion.

Answers

Answer:

45, 34, 36

Explanation:

The atomic number of Selenium is 34 and the atomic number is 79 also the atom has gained two electron denoted by superscript -2

number of neutrons = mass number - atomic number = 79 - 34 = 45

number of proton = atomic number = 34

number of electron = 34 + 2 = 36. In an atom the number of proton is always equal to number of electron if the atom is neutral but this Se atom has gain two so the number of electron will exceed the number of proton by 2.

The Se ion has 34 protons, 45 neutrons and 36 electrons.

The mass number (A) is given by the sum of the protons and neutrons:

A = protons + neutrons = 79

From the Periodic Table, we can see that the chemical element Selenium (Se) has an atomic number (Z) of 34, which is equal to the number of protons of a chemical element:

Z = protons = 34

Thus, we calculate the number of neutrons as the difference between A and Z:

neutrons = A - Z = 79 - 34 = 45

In a neutral atom (without electric charge), the number of electrons is equal to the number of protons. Since Se ion has 34 protons and a charge of -2, it has 34 electrons to be neutral and then it gained 2 electrons, so the number of electrons is equal to:

electrons = protons + 2 = 34 + 2 = 36

You can learn more about mass number, atomic number and subatomic particles here:

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The radius of an atom of gold (Au) is about 1.35 Å.How many gold atoms would have to be lined up to span 5.5 mm ?

Answers

The number of gold atoms that would be needed to span this distance is 20,370.4 atoms.

How many gold atoms would have to be lined up?

To calculate how many gold atoms would need to be lined up to span a given distance, we will us the following method.

The number of gold atoms that would be needed to span this distance:

Distance  = Diameter of a gold atom

Distance = 2 x Radius

Distance = 2 x 1.35 Å

Number of gold atoms = Total distance / Distance spanned by a single atom

Number of gold atoms = (5.5 x 10⁻⁴ cm) / (2 x 1.35 Å)

1 Å = 10⁻⁸ cm.

Number of gold atoms = (5.5 x 10⁻⁴ cm) / (2 x 1.35 x 10⁻⁸ cm)

Number of gold atoms = 20,370.4 atoms

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depends how many sig figs you are rounding to so I won't round to tenth or hundredth but the answer is 2.037 * 10^7

Calculate the pH of a buffer solution made by adding 15.0 g anhydrous sodium acetate (NaC2H3O2) to 100.0 mL of 0.200 M acetic acid. Assume there is no change in volume on adding the salt to the acid. (pKa for acetic acid is 4.74 or Ka is 1.8 x 10-5)3.

Answers

Answer:

pH of Buffer Solution 5.69

Explanation:

Mole of anhydrous sodium acetate = (Given mass)/(Molecular mass)

                                                           = (15)/(82)

                                                           = 0.18 mole

 100 ml of 0.2 molar acetic acid  means

= M x V

= 0.2 x 100

= 20 mmol

= 0.02 mole

Using Henderson equation to find pH of Buffer solution

pH = pKa + log([Salt])/([Acid])

     = 4.74 + log(0.18)/(0.02)

     = 4.74 + log 9

     = 5.69

So pH of the Buffer solution = 5.69