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alisha [4.7K]
3 years ago
9

Which of the following is the best example of a closed system?

Chemistry
2 answers:
xxMikexx [17]3 years ago
5 0
A greenhouse 
all others are wrong
Marat540 [252]3 years ago
3 0
Based on my information, this would be known to be a "green house" as this, would be as it would be a house known for planting, and etc. But the main part that I wanted to consider would be the fact that it would also be a "closed system" as you would be able to open and close this kind of machine.

Your answer: a greenhouse
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One peanut M&M weighs approximately 2.33 g.
Sphinxa [80]

Answer:

There are 23076 peanut M&M's in 53.768 kg of M&M's.

Explanation:

First we <u>convert 53.768 kg into g</u>:

  • 53.768 kg * 1000 = 53768 g

Then we <u>divide the total mass of M&M's by the mass of one peanut M&M,</u> in order to calculate the answer:

  • 53768 g / 2.33 g = 23076

So there are 23076 peanut M&M's in 53.768 kg of M&M's.

3 0
3 years ago
In general, non-spontaneous reactions __________.
Rom4ik [11]
Very low gives free energy
3 0
3 years ago
How many Helium atoms are in 12 grams of Helium?
horsena [70]
12gHe/1 × 1molHe/4.0026g × 6.02x10^23atomHe/1mol = 1.8 atoms
6 0
4 years ago
Why is secondary growth important to plants
stepan [7]
Secondary growth is important to plants because it involves thickening of the plant axis.It also increased amounts of vascular tissue.

I tried sorry if it’s not worded perfect :)
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
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