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Bad White [126]
3 years ago
11

At 1:00 P.M., the air temperature was 78°F, and air pressure was 30.0 inches of mercury with partly cloudy skies. By 4:00 P.M.,

the temperature dropped to 62°F, and the pressure dropped to 29.5 inches of mercury. Which best predicts the weather for later that afternoon? pls help im about to get killed by my teacher giving out brainiest to fastest answer THAT GOES WITH DA QUESTIONNN
Chemistry
2 answers:
bonufazy [111]3 years ago
7 0
I suppose it would be rain shower :)
vlabodo [156]3 years ago
5 0
The 4:00 time because later that afternoon would be later in the day
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Consider the reaction 2N2(g) O2(g)2N2O(g) Using the standard thermodynamic data in the tables linked above, calculate Grxn for t
ratelena [41]

Answer:

\Delta G^0 _{rxn} = 207.6\ kJ/mol

ΔG ≅ 199.91 kJ

Explanation:

Consider the reaction:

2N_{2(g)} + O_{2(g)} \to 2N_2O_{(g)}

temperature = 298.15K

pressure = 22.20 mmHg

From, The standard Thermodynamic Tables; the following data were obtained

\Delta G_f^0  \ \ \ N_2O_{(g)} = 103 .8  \ kJ/mol

\Delta G_f^0  \ \ \ N_2{(g)} =0 \ kJ/mol

\Delta G_f^0  \ \ \ O_2{(g)} =0 \ kJ/mol

\Delta G^0 _{rxn} = 2 \times \Delta G_f^0  \ N_2O_{(g)} - ( 2 \times  \Delta G_f^0  \ N_2{(g)} +   \Delta G_f^0  \ O_{2(g)})

\Delta G^0 _{rxn} = 2 \times 103.8 \ kJ/mol - ( 2 \times  0 +   0)

\Delta G^0 _{rxn} = 207.6\ kJ/mol

The equilibrium constant determined from the partial pressure denoted as K_p can be expressed as :

K_p = \dfrac{(22.20)^2}{(22.20)^2 \times (22.20)}

K_p = \dfrac{1}{ (22.20)}

K_p = 0.045

\Delta G = \Delta G^0 _{rxn} + RT \ lnK

where;

R = gas constant = 8.314 × 10⁻³ kJ

\Delta G =207.6 + 8.314 \times 10 ^{-3} \times 298.15  \ ln(0.045)

\Delta G =207.6 + 2.4788191 \times \ ln(0.045)

\Delta G =207.6+ (-7.687048037)

\Delta G = 199.912952  kJ

ΔG ≅ 199.91 kJ

7 0
4 years ago
How do I solve this?
ira [324]
Sorry, I can't really see the question )-:
6 0
4 years ago
PLEASE HELP!!! ON A TIMER, WILL MARK BRAINLIEST
Nadusha1986 [10]

Answer:

<em>Well, Your best answer will be is  2H+ + 2OH-  ->  2H2O but you have to reduce it to H+ + OH-  ->  H2O. </em><em>Good Luck!</em>

8 0
3 years ago
Read 2 more answers
A sample of gas is inside of a rigid container with fixed volume of 350mL. The initial pressure of the system is 366kPa, and the
Cerrena [4.2K]

Answer:

457.5kPa

Explanation:

Given data

V1=V2=350mL  (<em>fixed volume</em> )

P1=366kPa

T1= 88 degrees Celsius

P2=??

T2= 110 degrees Celsius

For the general gas equation

P1V1/T1= P2V2/T2

V1=V2

P1/T1= P2/T2

Substitute

366/88= P2/110

Cross multiply we have

P2*88=366*110

P2*88= 40260

P2= 40260/88

P2= 457.5 kPa

Hence the pressure will change to 457.5kPa

3 0
3 years ago
What is the value of the σ∗1s mo wave function at the nodal plane?
Leona [35]

Answer:

There is None

Explanation:

This is because it is a derived function dependent on other factors.

5 0
3 years ago
Read 2 more answers
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