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2 years ago

25 POINT QUESTION! HELP ASAP! WILL GIVE BRAINLIEST TO CORRECT ANSWER!!

Chemistry

1 answer:

Mice21 [21]2 years ago

5 0

Answer:

Explanation:

A catalyst functions to speed up chemical reaction. It does this by lowering the activation energy required for the reaction to begin by providing an alternative pathway for the substrates to interact with each.

The pathway ABD involves a reaction that requires a higher activation energy as compared to reaction following the pathway ACD as seen in the diagram.

So it means that a catalyst must be mediating the reaction with the lower activation energy. -pathway ACD, and this reaction will proceed faster than the reaction following the pathway ABD.

Hence option D is the only valid and correct option.

<h3>Hope this helps!</h3>

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Identify the limiting reactant in the reaction of nitrogen and hydrogen to form NH3 if 5.23 g of N2 and 5.52 g of H2 are combine

Marrrta [24]

Answer:

1) The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.

2) The amount (in grams) of excess reactant H₂ = 4.39 g.

Explanation:

Firstly, we should write the balanced equation of the reaction:

N₂ + 3H₂ → 2NH₃.

1) To determine the limiting reactant of the reaction:

From the stichiometry of the balanced equation, 1.0 mole of N₂ reacts with 3.0 moles of H₂ to produce 2.0 moles of NH₃.

This means that N₂ reacts with H₂ with a ratio of (1:3).

We need to calculate the no. of moles (n) of N₂ (5.23 g) and H₂ (5.52 g) using the relation: n = mass / molar mass.

The no. of moles of N₂ in (5.23 g) = mass / molar mass = (5.23 g) / (28.00 g/mol) = 0.1868 mol.

The no. of moles of H₂ (5.52 g) = mass / molar mass = (5.52 g) / (2.015 g/mol) = 2.74 mol.

From the stichiometry, N₂ reacts with H₂ with a ratio of (1:3).

The ratio of the reactants of N₂ (5.23 g, 0.1868 mol) to H₂ (5.52 g, 2.74 mol) is (1:14.67).

∴ The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.

0.1868 mol of N₂ react completely with 0.5604 mol of H₂ and the remaining of H₂ is in excess.

2) To determine the amount (in grams) of excess reactant of the reaction:

As showed in the part 1, The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.

Also, 0.1868 mol of N₂ react completely with 0.5604 mol of H₂ and the remaining of H₂ is in excess.

The no. of moles are in excess of H₂ = 2.74 mol - 0.5604 mol (reacted with N₂) = 2.1796 mol.

∴ The amount (in grams) of excess reactant H₂ = n (excess moles) x molar mass = (2.1796 mol)((2.015 g/mol) = 4.39 g.

4 0

2 years ago

Two products of stars

Andre45 [30]

Two properties of a star is luminosity and temperature.

4 0

3 years ago

Someone please help! Hurry

makkiz [27]

It’s b) Phototropism

4 0

2 years ago

Read 2 more answers

What is the mass, in pounds, of 389 mL of a gas that has a density of 1.29 g/L?

devlian [24]

Answer: Mass of gas is 0.001 pounds.

Explanation:

Density is defined as the mass contained per unit volume.

$Density=\frac{mass}{Volume}$

Given : Mass of gas = ?

Density of gas = $1.29g/L$

Volume of gas = 389 ml = 0.389 L (1L=1000ml)

Putting in the values we get:

$1.29g/L=\frac{mass}{0.389L}$

$mass=0.5grams$

$mass=0.5\times 0.002lb=0.001lb$ (1g =0.002 lb)

Thus the mass of gas is 0.001 pounds.

7 0

2 years ago

PLEASE HELP MEEE!!!!

Assoli18 [71]

It would be 35.8 Calories or calories. Not sure about that part. Hope this helps though.

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2 years ago