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snow_lady [41]
3 years ago
15

Relate pressure to the area over which a force is applied.

Chemistry
2 answers:
Alinara [238K]3 years ago
7 0
Altitude and oxygen
In-s [12.5K]3 years ago
5 0
I agree with the person above me
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What is the 2s in 42,256
MissTica

42,256 = 2,000

42,256 = 200

together they'd be 2,200 (if that's what you needed as well)

5 0
4 years ago
Read 2 more answers
NEED HELP PLSS:(
stira [4]

Answer:

A: 92.02 g

Explanation:

Equation of the reaction;

N2 (g)+ 2O2(g)------> 2NO2(g)

Note that the balanced reaction equation is the first step in solving any problem on stoichiometry. Once the reaction equation is correct, the question can be easily solved...

6 0
2 years ago
For the reaction 2Co3+(aq)+2Cl−(aq)→2Co2+(aq)+Cl2(g). E∘=0.483 V2Co3+(aq)+2Cl−(aq)→2Co2+(aq)+Cl2(g). E∘=0.483 V what is the cell
frutty [35]

Answer:

\large\boxed{\text{0.512 V}}

Explanation:

We must use the Nernst equation

E = E^{\circ} - \dfrac{RT}{zF}lnQ

The equation for the cell reaction is is

2Cl⁻(0.384 mol·L⁻¹) + 2Co³⁺(0.324 mol·L⁻¹) ⇌ Cl₂(5.80 atm) + 2Co²⁺(0.158 mol/L)

Data:

E° = 0.483 V

R = 8.314 J·K⁻¹mol⁻¹

T = 25 °C

n = 2

F = 96 485 C/mol

 

Calculation:  

T = 25 + 273.15 = 298.15 K

Q = \dfrac{\text{[Cl}^{-}]^{2}[\text{Co}^{3+}]^{2}}{p_{\text{Cl}_{2}}^{2}\text{[Co}^{3+}]^{2}} = \dfrac{0.384^{2} \times 0.324^{2}}{5.80 \times 0.158^{2}} =0.1069\\\\E = 0.483 - \left (\dfrac{8.314 \times 298.15 }{2 \times 96485}\right ) \ln(0.1069)\\\\=0.483 -0.01285 \times (-2.236) = 0.483 + 0.02872 = \textbf{0.512 V}\\\text{The cell potential is } \large\boxed{\textbf{0.512 V}}

7 0
4 years ago
Draw the structures of organic compounds a and<br> b. omit all byproducts
Georgia [21]

81. There is 1 carbon, 2 chlorine and fluorine atoms in Freon 12. To draw them it forms a cross with C in the middle and Cl and F both on the opposite side. 
     


8 0
3 years ago
The cell wall in plants functions to prevent the cell membrane from breaking when too much water enters the cell. When too much
Gnoma [55]

Answer:

<h2>                                   YOUR WELCOME</h2>
8 0
4 years ago
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