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3 years ago

An inherited behavior that helps an animal meet its needs

Chemistry

1 answer:

Sergeeva-Olga [200]3 years ago

4 0

Answer:

the answer is adaptations

Explanation:

A body part or behavior that helps an animal meet its need in its environment is called an adaptation. All adaptions are inherited traits. Camouflage is an animal's color or pattern that help it blend in with its surroundings.

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For the following reaction, 25.2 grams of sulfur dioxide are allowed to react with 5.36 grams of oxygen gas. sulfur dioxide (g)

Marizza181 [45]

Answer:

Maximum amount of sulfur trioxide that can be formed = 26.822 g

Explanation:

The balanced chemical equation for the reaction

2SO₂ + O₂ -----> 2SO₃

25.2 grams of sulfur dioxide are allowed to react with 5.36 grams of oxygen gas. What is the maximum amount of sulfur trioxide that can be formed?

It is the limiting reagent (the reactant in the stoichiometric lesser amount) that determines how much product is formed or how much of the other reactant is formed.

So, we convert the masses of reactants present into number of moles to get a clearer picture.

(Number of moles) = (mass)/(molar mass)

For sulfur dioxide,

Mass present = 25.2 g

Molar mass = 64.066 g/mol

(Number of moles present) = (25.2/64.066)

(Number of moles present) = 0.39 moles

For Oxygen gas,

Mass present = 5.36 g

Molar mass = 32.0 g/mol

(Number of moles present) = (5.36/32)

(Number of moles present) = 0.1675 moles

But from the stoichiometric balance,

2SO₂ + O₂ -----> 2SO₃

2 moles of Sulfur dioxide reacts with 1 mole of Oxygen gas

If Sulfur dioxide was the limiting reagent,

0.39 moles would react with (0.39×1/2) moles of Oxygen gas; 0.195 moles of Oxygen gas.

This is more than the total amount of Oxygen gas present at the start of the reaction, hence, Sulfur dioxide cannot be the limiting reagent.

Oxygen gas as limiting reagent,

1 mole of Oxygen gas reacts with 2 moles of Sulfur dioxide,

0.1675 moles of Oxygen gas would react with (0.1675×2/1) of Sulfur dioxide; 0.335 moles of Sulfur dioxide.

This indicates that oxygen is truly the limiting reagent and Sulfur dioxide is the reagent that is present in excess.

So, now, we calculate the amount of Sulfur trioxide that can be obtained from this reaction setup (assuming a 100% conversion and the maximum amount of Sulfur dioxide formed)

2SO₂ + O₂ -----> 2SO₃

1 mole of Oxygen gas gives 2 moles of Sulfur trioxide,

0.1675 moles of Oxygen gas will give (0.1675×2/1) moles of Sulfur trioxide; 0.335 moles of Sulfur trioxide.

We then convert this to mass.

(Mass) = (number of moles) × (molar mass)

Molar mass of SO₃ = 80.066 g/mol

(Mass of SO₃ produced) = 0.335 × 80.066

(Mass of SO₃ produced) = 26.822 g.

Maximum amount of sulfur trioxide that can be formed = 26.822 g

Hope this helps!!

5 0

3 years ago

Read 2 more answers

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

What is the molar mass of Ca(OH)2?

liubo4ka [24]

Answer: the molar mass is 74.092 g/mol

Explanation:

8 0

3 years ago

Water from a lake changes to a gas state in the process of what?

boyakko [2]

Of heating. Or when the lake is exposed to boil because of the temperature.

7 0

3 years ago

Read 2 more answers

.One ballon is filled with Hydrogen (H2) gas, the other balloon is filled with methane (CH4). Both balloons have the same temper

Neko [114]

Answer:

C. The balloon with CH4 has the same moles of gas molecules as the balloon with H2

Explanation:

Based on combined gas law, gases under the same pressure, temperature and volume have the same number of moles. With this information we can say the rigth statement is:

<h3>C. The balloon with CH4 has the same moles of gas molecules as the balloon with H2</h3>

7 0

3 years ago