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Dafna1 [17]
3 years ago
6

Two trials were conducted with these conditions:

Chemistry
1 answer:
alexandr402 [8]3 years ago
4 0
C or a i really cant be sure i need brainlest plz
most likely a
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_2. The temperature of a 100. mL sample of gas is increased while the volume it occupies decreases. The
yaroslaw [1]

Answer

the answer is b bro

Explanation:

i did my research and got it right

5 0
3 years ago
Determine the free energy(ΔG) from the standard cell potential (Ecell0 ) for the reaction:2ClO2-(aq)+Cl2(g)→2ClO2(g)+ 2Cl-(aq)wh
Dima020 [189]

<u>Answer:</u> The \Delta G^o for the given reaction is -7.84\times 10^4J

<u>Explanation:</u>

For the given chemical reaction:

2ClO_2^-(aq.)+Cl_2(g)\rightarrow 2ClO_2(g)+2Cl^-(aq.)

Half reactions for the given cell follows:

<u>Oxidation half reaction:</u> ClO_2^-\rightarrow ClO_2+e^-;E^o_{ClO_2^-/ClO_2}=0.954V  ( × 2)

<u>Reduction half reaction:</u> Cl_2+2e^-\rightarrow 2Cl(g);E^o_{Cl_2/2Cl^-}=1.36V

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=1.36-(0.954)=0.406V

To calculate standard Gibbs free energy, we use the equation:

\Delta G^o=-nFE^o_{cell}

Where,

n = number of electrons transferred = 2

F = Faradays constant = 96500 C

E^o_{cell} = standard cell potential = 0.406 V

Putting values in above equation, we get:

\Delta G^o=-2\times 96500\times 0.406=-78358J=-7.84\times 10^4J

Hence, the \Delta G^o for the given reaction is -7.84\times 10^4J

4 0
3 years ago
Which of the following statements is true?
liubo4ka [24]

Answer:

a is the anwser

Explanation:

4 0
3 years ago
Al2O3 + Fe Yields Fe3O4 + Al need before tomorrow
ella [17]
=<span><span><span><span><span><span><span><span><span>148.413159d</span><span>f2</span></span>i</span>l</span><span>o4</span></span>s</span>y</span>+<span><span>a<span>l2</span></span><span>o3</span></span></span>+<span>a<span>l is the answer and have a nice day :)</span></span></span>
4 0
3 years ago
The compound 1-iodododecane is a nonvolatile liquid with a density of 1.20g/ml. the density of mercury is 13.6g/ml. part a what
valina [46]

As the atmospheric pressure is, P = dgh

Here d is the density of the mercury,

g is gravitation = 9.8 m/s²

h is height of the column, P = 751 torr = (751 torr × 1 atm / 760 torr) (101325 Pa) (1 N/m² / 1 Pa) = 100125 N/m²

Where, 1 N = 1 Kg / ms²

Thus, P = 100125 Kg / m³. s²

Therefore, height of the mercury column, when the atmospheric pressure is 751 torr,

h = P / gd

= (100125 kg / m³. s²) / (9.8 m/s²) (13.6 × 10³ kg / m³) = 0.751 m

As, d₁h₁ = d₂h₂

Here, d₁ is the density of the non-volatile liquid = 1.20 g/ml

d₂ is the density of the mercury = 13.6 g/ml

h₂ = 0.751 m

Thus, putting the values we get,

h₁ = d₂h₂ /d₁ = 13.6 g/ml × 0.751 m / 1.20 g/ml

= 8.5 m


3 0
3 years ago
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