Which statement below is false when 10 g of nitrogen reacts with 5.0 g of hydrogen to produce ammonia?
N2 (g) + 3 H2 (g) → 2 NH3 (g)

Answers

Answer 1

Answer:

The theoretical yield of ammonia is 15 g.

Explanation:

The statements are:

2.8g of hydrogen are left over.

Hydrogen is the excess reactant.

Nitrogen is the limiting reactant.

The theoretical yield of ammonia is 15 g.

To solve this question we must find the moles of each reactant. Thus, we can find the limiting reactant. With limitinf reactant we can find the theoretical yield and the amount in excess of the reactant that is in excess:

Moles Nitrogen -28g/mol-

10.0g * (1mol / 28g) = 0.357 moles

Moles hydrogen -2g/mol-

5.0g * (1mol / 2g) = 2.5 moles

For a complete reaction of 0.357 moles of nitrogen are required:

0.357mol N2 * (3mol H2 / 1molN2) = 1.071 moles hydrogen.

As there are 2.5 moles of hydrogen, hydrogen is the excess reactant and nitrogen is the limiting reactant

The moles of hydrogen that are left over are:

2.5moles - 1.071moles = 1.429moles Hydrogen

The mass is:

1.429moles Hydrogen * (2g / mol) = 2.8g of hydrogen are left over

The moles of ammonia produced are:

0.357mol N2 * (2mol NH3 / 1molN2) = 0.714 moles of ammonia are produced (Theoretical moles)

The theoretical mass is (Molar mass NH3 = 17g/mol):

0.714 moles of ammonia * (17g / mol) =

12g of ammonia is the theoretical mass

The false statement is:

The theoretical yield of ammonia is 15 g.

Answer 2

The false statement, when 10 g of nitrogen reacts with 5.0 g of hydrogen to produce ammonia is that the theoretical yield of ammonia is 15g.

How moles is calculated?

Moles of any substance will be calculated as:

n = W/M, where

W = given mass

M = molar mass

Given chemical reaction is:
N₂(g) + 3 H₂(g) → 2 NH₃(g)

Moles of 10g of N₂ = 10/28 = 0.357 moles

Moles of 5g of H₂ = 5/2 = 2.5 moles

From the stoichiometry of the reaction:

1 mole of N₂ = react with 3 moles of H₂

0.357 moles of N₂ = react with 3×0.357=1.071 moles of H₂

Here nitrogen is the limiting reactant and hydrogen is excess reactant, and formation of ammonia depends on the limiting reactant.

1 mole of N₂ = react with 2 moles of NH₃

0.357 moles of N₂ = react with 2×0.357=0.714 moles of NH₃

Now we calculate the mass of ammonia from the given moles as:

M = (0.714) (17) = 12 g.

Hence, the false statement is that theoretical yield of ammonia is 15g.

To know more about moles, visit the below link:
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A balloon is filled with oxygen gas with a volume of 8.70 ML and a temperature of 318.2 K if the temperature increases to 337.8 K what would the new volume be

Answers

Answer:

The new volume will be 9.24 mL.

Explanation:

Charles's law determines that for a given sum of gas at constant pressure, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases.

So, Charles's law is a law that says that when the amount of gas and pressure are kept constant, the quotient that exists between the volume and the temperature will always have the same value:

[tex]\frac{V}{T} =k[/tex]

When studying an initial state 1 and a final state 2, it is satisfied:

[tex]\frac{V1}{T1} =\frac{V2}{T2}[/tex]

In this case:

V1= 8.70 mLT1= 318.2 KV2= ?T2= 337.8 K

Replacing:

[tex]\frac{8.70 mL}{318.2 K} =\frac{V2}{337.8 K}[/tex]

Solving:

[tex]V2=337.8K*\frac{8.70 mL}{318.2 K}[/tex]

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convert 555 torr into kpa

Answers

Answer:

73.97 kPa

Explanation:

It's helpful to know the conversion factors and apply them.

First, convert from torr to atm:

555 torr ÷ 760 atm = 0.730 atm

Next, convert from atm to Pa:

0.730 atm × 101,325 Pa = 73967 Pa

Finally, convert to kPa (1000 Pa in 1 kPa):

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If 0.583 g of ammonia (NH3) is dissolved to make 250 mL of solution, what is the molarity?

Answers

Answer:

0.137 M NH3

Explanation:

First divide the mass of NH3 by the molar mass of NH3, and then divide by the volume to get molarity.

0.583 g / 17.031 g/mol = 0.0342 mol NH3

0.0342 mol NH3 / 0.250 L = 0.137 M NH3

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Answers

Answer:

option D

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first quantum number of a 2s2

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second quantum number of a 2s2

l = 0 to n-1

l = 0, 1

For s orbital

l = 0

third quantum number of a 2s2

ml = - l to +l

ml = 0

fourth quantum number of a 2s2 electron be

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½ = spin up

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ms = +1/2

Answer:

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Answer:

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Answer:

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Answer:

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Answers

Answer:

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Explanation:

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Answer:

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Answers

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2. F

3. T

4. F

5. T

6. T

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Answers

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Answers

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please i need help thank you

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What he said is correct!!

HELPPP!!!
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0.70
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Answers

i think its C not so sure rho

good ,uck!!!!

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Rate law can be defined as a chemical equation that is typically used to relate the initial (forward) chemical reaction rate with respect to the concentrations or pressures of the chemical reactants and constant parameters.

Mathematically, the rate law is given by this formula:

[tex]R = K[A]^x[B]^y[C]^z[/tex]

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Answer:

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"Barium hydroxide forms a strong caustic base in aqueous solution. It has many uses, e.g., as a test for sulphides; in pesticides; in the manufacture of alkali and glass. Use of barium hydroxide lime rather than soda lime, high sevoflurane concentration, high absorbent temperature, and fresh absorbent use."

==================================================================

Hope I Helped, Feel free to ask any questions to clarify :)

Have a great day!

More Love, More Peace, Less Hate.

    -Aadi x

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h

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gy

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HELPPPPPPPPPPPPPPPPP

Answers

The water is formed from oxygen gas and...hydrogen gas, I'm assuming? It would have been nice for the question to have been a bit more explicit (not blaming you, of course).

Assuming that's the case, our chemical reaction would be:

2H₂(g) + O₂(g) → 2H₂O(l).

We are told that 1 mol of a gas has a volume of 24.0 dm³ at RTP. We can use this relation to determine the number of moles of O₂ gas that reacts given its initial volume, 33.5 dm³.

33.5 dm³ O₂(g)/24.0 dm³/mol = 1.396 mol O₂(g).

Since we are not given any information about H₂(g), or any other reactant for that matter, I am assuming that the O₂(g) is the limiting reactant. According to the equation, the stoichiometric ratio between O₂ and H₂O is 1:2. That is, for every one mole of O₂ that is consumed, two moles of H₂O are formed (i.e., the number of moles of H₂O formed is double the number moles of O₂).

Since 1.396 moles of O₂ reacts, 2(1.396) = 2.792 moles of H₂O are produced. To convert moles of water to grams, we multiply the number of moles of H₂O by the molar mass of H₂O:

(2.792 moles H₂O)(18.015 g/mol) = 50.3 g H₂O.

So, approximately 50.3 grams of water are formed from 33.5 dm³ of oxygen gas at RTP.

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