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Ket [755]
2 years ago
9

How many mm are in 38,987 L?

Chemistry
1 answer:
OlgaM077 [116]2 years ago
6 0

Answer:

38987000

Explanation:

there are 1000 ml in a Liter.

Hope this helps!

You might be interested in
How many grams of oxygen are required to react with 13.0 grams of octane (C8H18) in the combustion of octane in gasoline?
egoroff_w [7]

Grams of oxygen are required to react with 13.0 grams of octane (C8H18) in the combustion of octane in gasoline is 45.5g

Octane is a hydrocarbon which burns in gasoline in presence of oxygen according to the given balanced chemical equation,

2C₈H₁₈ + 25O₂------> 16CO₂ + 18H₂0

Molar mass of octane = 114.23g/mol

Molar mass of Oxygen = 32g/mol

According to the stiochiometry of the balanced equation the mole ratio of Octane and Oxygen is 2:25

2 mole of octane needs 25 mole of oxygen

1 mole of octane needs 12.5 moleof oxygen

114.23g of octane needs 400g of oxygen

13g   of octane  needs 45.5g of oxygen

Mass of oxygen needed =45.5g

Hence, the Mass of oxygen needed is 45.5g for the combustion of octane in gasoline.

Learn more about Octane here, brainly.com/question/21268869

#SPJ4

5 0
1 year ago
If 60 grams of a liquid takes up 120ml how dense is the liquid
Black_prince [1.1K]

Answer:

<h2>Density = 0.5 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

<h3>Density =  \frac{mass}{volume}</h3>

From the question

mass = 60 g

volume = 120 mL

Substitute the values into the above formula and solve

That's

<h3>Density =  \frac{60}{120}  \\  =  \frac{1}{2}</h3>

We have the final answer as

<h3>Density = 0.5 g/mL</h3>

Hope this helps you

4 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
If two metals has the same specific heat how can distinguish them in two ways​
tensa zangetsu [6.8K]

Explanation:

1) Thermal and Electric conductivity

2) Metallic strength

5 0
2 years ago
Photosynthesis was another biological phenomenon that occupied the attention of the chemists of the late 18th century. The demon
balu736 [363]

Answer:

In the 1770s, the English clergyman Joseph Priestley (who is credited with the discovery of O2) established the production of oxygen by vegetables recognizing that the process was, apparently, the inverse of animal respiration, which consumed such chemical element.

Explanation:

In 1772, Joseph Priestley in his Recherches sur diversces especes d'air differentiated the air of animal respiration from that emitted by vegetables in the presence of light. Of the latter, which he called "dephlogistic air", he highlighted his purifying property of the environment indicating that: plants far from affecting the air in the same way as animal respiration, produce the opposite effects, and tend to preserve the sweet and healthy atmosphere , when it becomes harmful as a result of the life and breathing of the animals or their death and their rot.

In 1780, Jean Ingeshousz in his Experiences sur les vegetaux completed and reaffirmed the observations of Joseph Priestley. At the same time, he could deny Charles Bonnet's hypothesis, by demonstrating that the air expelled from the leaves comes from inside, and that the stimulating factor of the gaseous emission was not the heat produced by the sun, but the intensity of the light .

It was, finally, Jean Senebier that between 1782 and 1784, found that the "fixed air" dissolved in the water favors the vegetation. From these observations, he hypothesized that "fixed air" (carbon dioxide) is absorbed by the plants, which take it from the atmosphere with the humidity it has and in which it is mixed. Once this gas has been captured, both from the atmosphere and from the ground, it is decomposed in the presence of light by the leaves, releasing the "vital air" (oxygen) and leaving the carbon in the plant.

Thus, at the end of the century the participation of the atmosphere in plant dynamics was already seated, although the how and why of this participation were still unknown and no theory had been formulated to explain the nutritional process as a whole.

3 0
3 years ago
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