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4 years ago

How do producers and consumers get energy from the sun?

1 answer:

6 0

Answer: In photosynthesis, producers combine carbon dioxide, water, and sunlight to produce oxygen and sugar (their food). Other organisms get energy by eating producers. ... It cannot directly use the Sun's energy to make food. As a consumer, it has to eat— or, consume— other organisms for energy.

Explanation: Thats how both producers and consumers get energy

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A titration of 25.0 mL of a solution of the weak base aniline, C6H5NH2, requires 25.67 mL of 0.175 M HCl to reach the equivalenc

Lorico [155]

Answer:

a. 0.180M of C₆H₅NH₂

b. 0.0887M C₆H₅NH₃⁺

c. pH = 2.83

Explanation:

a. Based in the chemical equation:

C₆H₅NH₂(aq) + HCl(aq) → C₆H₅NH₃⁺(aq) + Cl⁻(aq)

<em>1 mole of aniline reacts per mole of HCl</em>

Moles required to reach equivalence point are:

Moles HCl = 0.02567L ₓ (0.175mol / L) = 4.492x10⁻³ moles HCl = moles C₆H₅NH₂

As the original solution had a volume of 25.0mL = 0.0250L:

4.492x10⁻³ moles C₆H₅NH₂ / 0.0250L = 0.180M of C₆H₅NH₂

b. At equivalence point, moles of C₆H₅NH₃⁺ are equal to initial moles of C₆H₅NH₂, that is 4.492x10⁻³ moles

But now, volume is 25.0mL + 25.67mL = 50.67mL = 0.05067L. Thus, molar concentration of C₆H₅NH₃⁺ is:

[C₆H₅NH₃⁺] = 4.492x10⁻³ moles / 0.05067L = 0.0887M C₆H₅NH₃⁺

c. At equivalence point you have just 0.0887M C₆H₅NH₃⁺ in solution. C₆H₅NH₃⁺ has as equilibrium in water:

C₆H₅NH₃⁺(aq) + H₂O(l) → C₆H₅NH₂ + H₃O⁺

Where Ka = Kw / Kb = 1x10⁻¹⁴ / 4.0x10⁻¹⁰ =

When the system reaches equilibrium, molar concentrations are:

[C₆H₅NH₃⁺] = 0.0887M - X

[C₆H₅NH₂] = X

[H₃O⁺] = X

Replacing in Ka formula:

2.5x10⁻⁵ = [X] [X] / [0.0887M - X]

2.2175x10⁻⁶ - 2.5x10⁻⁵X = X²

0 = X² + 2.5x10⁻⁵X - 2.2175x10⁻⁶

Solving for X:

X = -0.0015 → False solution. There is no negative concentrations.

X = 0.001477 → Right solution.

As [H₃O⁺] = X, [H₃O⁺] = 0.001477

Knowing pH = -log [H₃O⁺]

pH = -log 0.001477

7 0

3 years ago

Erythrogenic acid, C18H26O2, is an acetylenic fatty acid that turns avivid red on exposure to light. On catalytic hydrogenation

lara31 [8.8K]

Answer:

use research

Explanation:

6 0

3 years ago

When you push or pull on something, you are using a

bearhunter [10]

The answer would be Force

6 0

3 years ago

Read 2 more answers

How mant grams are there in 7.15 x 10^24 molecules of water?

worty [1.4K]

Answer:

B. 214.02

Explanation:

1 mol of water weighs 18.015 gm and contains 6.023 × 10²³ molecules

From question, We have 7.15 × 10²⁴ molecules

Dividing we get (7.15 × 10 ²⁴) ÷ ( 6.023 × 10²³) = 11.871 molecules

Now, Weight of water = 11.871 × 18.015 = 213.85 which is nearer to option B

4 0

3 years ago

Read 2 more answers

1. Ethylene glycol, commonly used in antifreezes, contains only carbon, hydrogen, and oxygen. When a sample of it is combusted i

11111nata11111 [884]

Answer:

Empirical formula = CH3O

Molecular formula = C2H6O2

Explanation:

Step 1: Data given

Mass of the sample = 23.46 grams

Mass of H2O = 20.42 grams

Molar mass of H2O = 18.02 g/mol

Mass of CO2 = 33.27 grams

Molar mass of CO2 = 44.01 G:mol

Atomic mass of C = 12.01 g/mol

Atomic mass of O = 16.0 g/mol

Atomic mass of H = 1.01 g/mol

Molar mass of the compound = 62.0 g/mol

Step 2: Calculate moles of H2O

Moles H2O = 20.42 grams / 18.02 g/mol

Moles H2O = 1.133 moles

Step 3: Calculate moles H

For 1 mol H2O we have 2 moles H

For 1.133 moles H2O we have 2* 1.133 = 2.266 moles H

Step 4: Calculate mass H

Mass H = 2.266 moles * 1.01 g/mol

Mass H = 2.29 grams

Step 5: Calculate moles CO2

Moles CO2 = 33.27 grams / 44.01 g/mol

Moles CO2 = 0.7560 moles

Step 6: Calculate moles C

For 1 mol CO2 we have 1 mol C

For 0.7560 moles CO2 we have 0.7560 moles C

Step 7: Calculate mass C

Mass C = 0.7560 moles * 12.01 g/mol

Mass C = 9.08 grams

Step 8: Calculate mass O

Mass O = 23.46 grams - 9.08 grams - 2.29 grams

Mass O =12.09 grams

Step 9: Calculate moles O

Moles O = 12.09 grams / 16.0 g/moles

Moles O = 0.7556

Step 10: Calculate mol ratio

We divide by the smallest amount of moles

C: 0.7560 moles / 0.7556 moles =1

H: 2.266 moles / 0.7556 moles =3

O; 0.7556 / 0.7560 moles = 1

This means for 1 mol C we have 3 moles H and 1 mol O

The empirical formula is CH3O

Step 11: Calculate the molecular formula

The molar mass of the empirical formula is 31 g/mol

Step 11: Calculate molecular formula

We have to multiply the empirical formula by n

n = 62.0 g/mol / 31g/mol = 2

Molecular formula = 2*(CH3O)

Molecular formula = C2H6O2

5 0

3 years ago