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ivolga24 [154]
3 years ago
8

The first organism in a food chain muust always be what type e of organism

Chemistry
1 answer:
mel-nik [20]3 years ago
7 0
The first organisms are autotrophs (primary producers)
You might be interested in
When 500.0 g of water is decomposed by electrolysis and the yield of hydrogen is only 75.3%, how much hydrogen chloride can be m
Evgen [1.6K]

The amount of hydrogen chloride that can be made is 1064 g

Why?

The two reactions are:

2H₂O → 2H₂ + O₂ 75.3 % yield

H₂ + Cl₂ → 2HCl 69.8% yield

We have to apply a big conversion factor to go from grams of water (The limiting reactant), to grams of HCl, the final product. We have to be very careful with the coefficients and percentage yields!

500.0gH_2O*\frac{1moleH_2O}{18.01 gH_2O}*\frac{2 moles H_2}{2 moles H_2O}*\frac{2.015g H_2}{1 mole H_2}*\frac{75.3 actual g}{100 theoretical g}=42.12 g H_2

42.12H_2*\frac{1 mole H_2}{2.015gH_2}*\frac{2 moles HCl}{1 mole H_2}*\frac{36.46g}{1 mole HCl}*\frac{69.8 actualg}{100 theoreticalg} =1064gHCl

Have a nice day!

#LearnwithBrainly

7 0
3 years ago
CH3OH can be synthesized by the reaction:
Leviafan [203]

Answer:

5

Explanation:

8 0
3 years ago
Determine the pH of the resulting solution if 25 mL of 0.400 M strychnine (C21H22N2O2) is added to 50 mL of 0.200 M HCl? Assume
DIA [1.3K]

Answer:

pH = 4.56

Explanation:

The strychnine reacts with HCl as follows:

C₂₁H₂₂N₂O₂ + HCl ⇄ C₂₁H₂₂N₂O₂H⁺ + Cl⁻

<em />

For strychnine buffer:

pOH = 5.74 + log [C₂₁H₂₂N₂O₂H⁺] / [C₂₁H₂₂N₂O₂]

Initial moles of C₂₁H₂₂N₂O₂ are:

0.025L * (0.400 mol / L) = 0.01 moles C₂₁H₂₂N₂O₂

And of HCl are:

0.05L * (0.200 mol / L) = 0.01 moles HCl

That means after the reaction, you will have just 0.01 moles of C₂₁H₂₂N₂O₂H⁺ in 50mL + 25mL = 0.075L. And molarity is:

[C₂₁H₂₂N₂O₂H⁺] = 0.01 mol / 0.075L = 0.1333M

This conjugate acid, is in equilibrium with water as follows:

C₂₁H₂₂N₂O₂H⁺(aq) + H₂O(l) ⇄ C₂₁H₂₂N₂O₂ + H₃O⁺

<em />

<em>Where Ka = Kw / Kb = 1x10⁻¹⁴ / 1.8x10⁻⁶ = 5.556x10⁻⁹</em>

<em />

Ka is defined as:

Ka = 5.556x10⁻⁹ = [C₂₁H₂₂N₂O₂] [H₃O⁺] / [C₂₁H₂₂N₂O₂H⁺]

In equilibrium, concentrations are:

C₂₁H₂₂N₂O₂ = X

H₃O⁺ = X

C₂₁H₂₂N₂O₂H⁺ = 0.1333M - X

Replacing in Ka expression:

5.556x10⁻⁹ = [X] [X] / [0.1333M - X]

7.39x10⁻¹⁰ - 5.556x10⁻⁹X = X²

7.39x10⁻¹⁰ - 5.556x10⁻⁹X - X² = 0

Solving for X:

X = - 2.72x10⁻⁵M → False solution. There is no negative concentrations

X = 2.72x10⁻⁵M → Right solution.

As H₃O⁺ = X

H₃O⁺ = 2.72x10⁻⁵M

And pH = -log H₃O⁺

<h3>pH = 4.56</h3>
4 0
3 years ago
Which statement about atoms and molecules is
Pavel [41]

Elements always exist as pair of atoms called molecules .

Explanation:-

  • The material which has only one types of similar atoms called element .
  • Ex:-Sodium,Carbon etc
8 0
3 years ago
Read 2 more answers
What are the symbols for the elements with the following valence electron configurations? List all.a. s2d1b. s2p3c. s2p
maria [59]

Answer:

Sc : 4s 2 3d 1

 Y : l 5s 2 4d 1

La : 6s 2 5d 1

Ce : 6s 2 4f 1 5d 1

Gd : 6s 2 4f 7 5d 1

Lu : 6s 2 4f 14 5d 1

Ac : 7s 2 6d 1

Pa : 7s 2 5f 2 6d 1

U : l 7s 2 5f 3 6d 1

Np : 7s 2 5f 4 6d 1

Cm : 7s 2 5f 7 6d 1

b. s2p3

he pnictogens:

N l : 2s 2 2p 3

P l : 3s 2 3p 3

As : 4s 2 3d 10 4p 3

Sb : 5s 2 4d 10 5p 3

Bi : 6s 2 4f 14 5d 10 6p 3

Mc : 7s 2 5f 14 6d 10 7p 3

c.The noble gases:

Ne : 2s 2 2p 6

Ar : 3s 2 3p 6

Kr : 4s 2 3d 10 4p 6

Xe : 5s 2 4d 10 5p 6

Rn : 6s 2 4f 14 5d 10 6p 6

Og : 7s 2 5f 14 6d 10 7p 6

Explanation: From the periodic tables we can drive elements with the electronic configuration

Sc : 4s 2 3d 1

 Y : l 5s 2 4d 1

La : 6s 2 5d 1

Ce : 6s 2 4f 1 5d 1

Gd : 6s 2 4f 7 5d 1

Lu : 6s 2 4f 14 5d 1

Ac : 7s 2 6d 1

Pa : 7s 2 5f 2 6d 1

U : l 7s 2 5f 3 6d 1

Np : 7s 2 5f 4 6d 1

Cm : 7s 2 5f 7 6d 1

b. s2p3

he pnictogens:

N l : 2s 2 2p 3

P l : 3s 2 3p 3

As : 4s 2 3d 10 4p 3

Sb : 5s 2 4d 10 5p 3

Bi : 6s 2 4f 14 5d 10 6p 3

Mc : 7s 2 5f 14 6d 10 7p 3

c.The noble gases:

Ne : 2s 2 2p 6

Ar : 3s 2 3p 6

Kr : 4s 2 3d 10 4p 6

Xe : 5s 2 4d 10 5p 6

Rn : 6s 2 4f 14 5d 10 6p 6

Og : 7s 2 5f 14 6d 10 7p 6

3 0
3 years ago
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