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dalvyx [7]
3 years ago
10

If atmospheric pressure measurement for taking at regular intervals from sea level to the stratopause The measurements with most

likely show that the pressure
Chemistry
2 answers:
slava [35]3 years ago
6 0

Answer: Pressure decreases.

Explanation:

Atmospheric pressure is termed as the  downward force applied by the weight of air against a particular area on Earth's surface.  the Standard atmospheric pressure is determined  by weighing the column of air occupying  1 square inch on Earth's surface which begins  at sea level and moves  upward to the uppermost layer of the atmosphere

Measuring the Atmospheric pressure from sea level to the stratosphere,the second layer of the atmosphere which contains the most of the ozone gas show that the pressure decreases

Lapatulllka [165]3 years ago
5 0

Answer:

Troposphere

Explanation:

The troposphere is the lowest layer of Earth's atmosphere that extends from under the stratosphere down to the surface of the earth. Nearly all weather condition takes place in the troposphere. The troposphere contains about 75% of the atmosphere's mass and 99% of the total mass of water vapour and aerosols in the atmosphere.

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Which of the following atoms has the highest electronegativity?
katrin [286]
A. Chlorine is the most electronegative element out of the possible choices. Generally (though not 100% of the time) electronegativity increases as you go further left and up the periodic table.



3 0
3 years ago
2C3H6O + 8O2 --> 6CO2 + 6H2O
grigory [225]

Answer: 0.45 moles of H_2O will be produced from 0.15 moles of propanol.

Explanation:

The balanced chemical reaction is:

2C_3H_6O+8O_2\rightarrow 6CO_2+6H_2O

C_3H_6O is the limiting reagent as it limits the formation of product and O_2 is the excess reagent.

According to stoichiometry :

2 moles of C_3H_6Oproduces = 6 moles of H_2O

Thus 0.15 moles of V will produce=\frac{6}{2}\times 0.15=0.45moles  of H_2O

Thus 0.45 moles of H_2O will be produced from 0.15 moles of propanol.

7 0
3 years ago
In today’s experiment, some lab students let their chromatograms elute slightly longer than others. If one student’s solvent fro
larisa86 [58]

Answer:

They should obtain the same Rf for the same compounds.

Explanation:

The <em>Rf</em> is defined as A/B. Where A is the displacement of the substance of interest, and B is the solvent front.

By dividing the substance's displacement by B, we make it so that the Rf factor is equal for identical compounds in the same mobile phase, no matter what the solvent front is.

5 0
3 years ago
A particular reactant decomposes with a half‑life of 113 s when its initial concentration is 0.331 M. The same reactant decompos
algol13

Answer:

The reaction is second-order, and k = 0.0267 L mol^-1 s^-1

Explanation:

<u>Step 1:</u> Data given

The initial concentration is 0.331 M

half‑life time =  113 s

The same reactant decomposes with a half‑life of 243 s when its initial concentration is 0.154 M.

<u>Step 2: </u>Determine the order

The reaction is not first-order because the half-life of a first-order reaction is independent of the initial concentration:

t½ = (ln(2))/k

Calculate k for the two conditions given:

⇒ 113 s with initial concentration is 0.331 M

t½ = ([A]0)/2k

113 s = (0.331 M)/2k

k = 0.00146 mol L^-1 s^-1

⇒ 243 s with an initial concentration is 0.154 M

t½ = ([A]0)/2k

243 s = (0.154 M)/2k

k = 0.000317 mol L^-1 s^-1

The <u>values of k are different</u>, so that rules out zero-order.

<u>Step 3: </u>Calculate if it's a second-order reaction

For a second-order reaction, the half-life is given by the expression

t½ = 1/((k*)[A]0))

<u>Calculate k for the two conditions given: </u>

⇒ 113 s when its initial concentration is 0.331 M

t½ = 1/((k*)[A]0))

113 s = 1/(k*(0.331 M))

k = 1/((0.331 M)*(113 s)) = 0.0267 L mol^-1 s^-1

⇒ 243 s when its initial concentration is 0.154 M

t½ = 1/((k*)[A]0))

243 s = 1/(k*(0.154 M))

k = 1/((0.154 M)*(243 s)) =  0.0267 L mol^-1 s^-1

The values of k are the same, so the reaction is second-order, and k = 0.0267 L mol^-1 s^-1

4 0
4 years ago
A chemist fills a reaction vessel with 9.20 atm nitrogen monoxide (NO) gas, 9.15 atm chlorine (CI) gas, and 7.70 atm nitrosyl ch
ivanzaharov [21]

Answer:

The reactions free energy \Delta G = -49.36 kJ

Explanation:

From the question we are told that

      The pressure of (NO) is P_{NO} = 9.20 \ atm

      The  pressure of  (Cl) gas is  P_{Cl} = 9.15 \ atm

       The  pressure of nitrosly chloride (NOCl) is P_{(NOCl)} = 7.70 \ atm

The reaction is

              2NO_{(g)} + Cl_2 (g)    ⇆   2 NOCl_{(g)}

 From the reaction we can  mathematically evaluate the \Delta G^o (Standard state  free energy ) as

                    \Delta G^o = 2 \Delta G^o _{NOCl} -   \Delta G^o _{Cl_2}  - 2 \Delta G^o _{NO}

The Standard state  free energy for NO is  constant with a value  

                 \Delta G^o _{NO} = 86.55 kJ/mol

 The Standard state  free energy for Cl_2 is  constant with a value                  

             \Delta G^o _{Cl_2} = 0kJ/mol

 The Standard state  free energy for NOCl is  constant with a value

         \Delta G^o _{NOCl} =66.1kJ/mol

Now substituting this into the equation

        \Delta G^o = 2 * 66.1 - 0 - 2 * 87.6

                = -43 kJ/mol

The pressure constant is evaluated as

         Q =  \frac{Pressure \ of  \ product }{ Pressure  \ of \ reactant }

Substituting  values  

        Q = \frac{(7.7)^2 }{(9.2)^2 (9.15) } = \frac{59.29}{774.456}

           = 0.0765

The free energy for this reaction is evaluated as

           \Delta  G  =  \Delta  G^o  + RT ln Q

Where R is gas constant with a value  of  R = 8.314 J / K \cdot mol

          T is temperature in K  with a given value of  T = 25+273 = 298 K

   Substituting value

                \Delta  G  = -43 *10^{3} + 8.314 *298 * ln [0.0765]

                       = -43-6.36

                      \Delta G = -49.36 kJ

4 0
3 years ago
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