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larisa86 [58]
2 years ago
6

To completely convert 9. 0 moles of hydrogen gas (h2) to ammonia gas, how many moles of nitrogen gas (n2) are required?

Chemistry
1 answer:
evablogger [386]2 years ago
3 0

To completely convert 9. 0 moles of hydrogen gas (h2) to ammonia gas, 3.0 moles of nitrogen gas (n2) are required.

<h3>What are moles?</h3>

The mole is a SI unit of measurement that is used to calculate the quantity of any substance.

<h3 />

The given reaction is \rm  N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)

By the stoichiometry rule of ratio hydrogen: nitrogen

3 : 1

The reacted moles of nitrogen is equals to H/3 moles of reacted hydrogen

So, moles of nitrogen  

\rm Moles\; of\; nitrogen = \dfrac{9.0 }{3} =3.0\;mol

Thus, 3.0 moles of nitrogen gas (n2) are required.

Learn more about moles

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ziro4ka [17]

Answer:

Molecule

Explanation:

molecule of the substance.  You can break the molecule down further, into the atoms that make it up, but those don't have the properties of the original  'compound'.

Here's an example:

-- Sodium is a soft, slippery metal, that explodes when water touches it.

-- Chlorine is a poisonous green gas.

When an atom of Sodium and an atom of Chlorine combine, they make one molecule of a substance called "Sodium Chloride".  That's SALT !  It isn't green, it isn't a gas, it isn't poisonous, it isn't soft and slippery, and it doesn't explode when water touches it.

3 0
3 years ago
In the absence of sodium methoxide, the same alkyl bromide gives a different product. Draw an arrowpushing mechanism to account
hoa [83]

Answer:

See explanation below

Explanation:

The question is incomplete, cause you are not providing the structure. However, I found the question and it's attached in picture 1.

Now, according to this reaction and the product given, we can see that we have sustitution reaction. In the absence of sodium methoxide, the reaction it's no longer in basic medium, so the sustitution reaction that it's promoted here it's not an Sn2 reaction as part a), but instead a Sn1 reaction, and in this we can have the presence of carbocation. What happen here then?, well, the bromine leaves the molecule leaving a secondary carbocation there, but the neighbour carbon (The one in the cycle) has a more stable carbocation, so one atom of hydrogen from that carbon migrates to the carbon with the carbocation to stabilize that carbon, and the result is a tertiary carbocation. When this happens, the methanol can easily go there and form the product.

For question 6a, as it was stated before, the mechanism in that reaction is a Sn2, however, we can have conditions for an E2 reaction and form an alkene. This can be done, cause the extoxide can substract the atoms of hydrogens from either the carbon of the cycle or the terminal methyl of the molecule and will form two different products of elimination. The product formed in greater quantities will be the one where the negative charge is more stable, in this case, in the primary carbon of the methyl it's more stable there, so product 1 will be formed more (See picture 2)

For question 6b, same principle of 6a, when the hydrogen migrates to the 2nd carbocation to form a tertiary carbocation the methanol will promove an E1 reaction with the vecinal carbons and form two eliminations products. See picture 2 for mechanism of reaction.

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3 years ago
An experimental spacecraft consumes a special fuel at a rate of 372 L/min. The density of the fuel is 0.730 g/mL and the standar
Karolina [17]

Explanation:

First, we will calculate fuel consumption is as follows.

         372 L/min \times 1000 ml/L \times 0.730 g/ml \times \frac{1}{60} min/s

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Now, we will calculate the power as follows.

        Power = Fuel consumption rate × -enthalpy of combustion

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Thus, we can conclude that maximum power (in units of kilowatts) that can be produced by this spacecraft is 1.19 \times 10^{5} kW.

5 0
3 years ago
Is each of these statements true? If not, explain why.(m) For most reactions, ΔH°rnx is lowered by a catalyst.
barxatty [35]

A  catalyst reduces H°rnx in most reactions. The answer is false

<h3>Do catalysts reduce delta H?</h3>

By reducing the activation energy required for the reaction to occur, a catalyst just modifies the route used to go from reactants to products. However, because it doesn't alter the state of the products or reactants, delta H is unaffected.

A catalyst reduces a reaction's activation energy, enabling a chemical reaction to occur. The number of reactant particles with energy above the activation energy increases as the temperature of a reaction rises.

learn more about catalyst refer

brainly.com/question/12507566

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3 0
1 year ago
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