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KonstantinChe [14]
3 years ago
13

Fritz-Haber process

Chemistry
1 answer:
maks197457 [2]3 years ago
6 0

Answer:

5×10⁵ L of ammonia (NH3)

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

N2 + 3H2 —> 2NH3

From the balanced equation above, we can say that:

3 L of H2 reacted to produce 2 L of NH3.

Finally, we shall determine the volume of ammonia (NH3) produced by the reaction of 7.5×10⁵ L of H2. This can be obtained as illustrated below:

From the balanced equation above,

3 L of H2 reacted to produce 2 L of NH3.

Therefore, 7.5×10⁵ L of H2 will react to produce = (7.5×10⁵ × 2)/3 = 5×10⁵ L of NH3.

Thus, 5×10⁵ L of ammonia (NH3) is produced from the reaction.

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Using the temperature change provided, which of the following best describes the direction of energy flow in this reaction
kherson [118]

Answer:

: the answer is C

Explanation:

:State the law of conservation of energy.

Define and endothermic process.

Define an exothermic process.

Make conversions involving heat units.

7 0
2 years ago
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24) What is momentum? *
sergiy2304 [10]

Answer:

Momentum is the measure of the motion of an object found by multiplying the objects mass and velocity.

Symbol: p

Units: kg x m/s

Explanation:

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6 0
2 years ago
A 53. 5 mL ample of an 5. 4 % (m / v) KBr olution i diluted with water o that the final volume i 205. 0 mL Expre your anwer to t
gogolik [260]

The concentration after dilution is 1.4%.

We are aware that concentration and volume are related to each other by the formula -

C_{1} V_{1} = C_{2} V_{2}, where we have initial concentration and volume on Left Hand Side and final concentration and volume on Right Hand Side.

Keep the values to calculate final concentration.

C_{2} = (53.5 × 5.4)/205.0

Performing multiplication on Right and Side

C_{2} = 288.9/205.0

Performing division on Right Hand Side

C_{2} = 1.4%

Hence, the final concentration is 1.4%.

Learn more about concentration -

brainly.com/question/17206790

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The complete question is -

A 53.5 mL sample of an 5.4 % (m/v) KBr solution is diluted with water so that the final volume is 205.0 mL.

Calculate the final concentration and express your answer to two significant figures and include the appropriate units.

3 0
1 year ago
Concentrations-
Arisa [49]
Moles of HCl is 3.47mol

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6 0
3 years ago
Determine the mass of Al(C2H3O2)3 that contains 2.63 Χ 1024 atoms of oxygen.
sergejj [24]

The molecular formula of compound is Al(C_{2}H_{3}O_{2})_{3}.

Since, in 1 mole of  Al(C_{2}H_{3}O_{2})_{3} there are 6.023\times 10^{23} atoms of  Al(C_{2}H_{3}O_{2})_{3}.

Thus, according to molecular formula, in 1 mole of  Al(C_{2}H_{3}O_{2})_{3} there are 6\times 6.023\times 10^{23}=3.61\times 10^{24} atoms of oxygen atoms.

1 atom of oxygen will be present in \frac{1}{3.61\times 10^{24}} moles of Al(C_{2}H_{3}O_{2})_{3} . Thus,

2.63\times 10^{24} atoms of oxygen \rightarrow \frac{2.63\times 10^{23}}{3.61\times 10^{24}}= 0.7285 moles of Al(C_{2}H_{3}O_{2})_{3}.

Molar mass of Al(C_{2}H_{3}O_{2})_{3} is 236 g/mol, mass can be calculated as follows:

m=n\times M=0.7285 mol\times 236 g/mol=172 g

Therefore, mass of Al(C_{2}H_{3}O_{2})_{3} will be 172 g.

5 0
3 years ago
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