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Svet_ta [14]
3 years ago
14

Be sure to answer all parts.

Chemistry
1 answer:
Fudgin [204]3 years ago
7 0

Answer:

2H₂ + O₂       →     2H₂O

Explanation:

Chemical equation:

H₂ + O₂       →     H₂O

Balance chemical equation:

2H₂ + O₂       →     2H₂O

Step 1:

H₂ + O₂       →     H₂O

Left hand side                     Right hand side

H = 2                                    H = 2

O = 2                                    O = 1

Step 2:

H₂ + O₂       →     2H₂O

Left hand side                     Right hand side

H = 2                                    H = 4

O = 2                                    O = 2

Step 3:

2H₂ + O₂       →     2H₂O

Left hand side                     Right hand side

H = 4                                    H = 4

O = 2                                    O = 2

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Answer:

False

Explanation:

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For the reaction H2PO4- HAsO4 2-HPO4 2-+H2AsO4- <br> what species are a conjugate acid-base pair?
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Ionic equation:

\text{H}_2\text{PO}_4^{-} + \text{HAsO}_4^{2-} \to \text{HPO}_4^{2-} + \text{H}_2\text{AsO}_4^{-}

The acid and base in a conjugate pair differ by only one proton \text{H}^{+}. The acid loses one proton to produce a conjugate base, whereas the base gains a proton to produce its conjugate acid.

\text{H}_2\text{PO}_4^{-} loses one proton to produce \text{HPO}_4^{2-} in this reaction.

\text{H}_2\text{PO}_4^{-} \to \text{H}^{+} + \text{HPO}_4^{2-}

Meanwhile, \text{HAsO}_4^{2-} gains one proton to form \text{H}_2\text{AsO}_4^{-}.

\text{HAsO}_4^{2-} + \text{H}^{+} \to \text{H}_2\text{AsO}_4^{-}

Therefore

  • \text{H}_2\text{PO}_4^{-} is the conjugate acid  \text{HPO}_4^{2-}, its conjugate base.
  • \text{HAsO}_4^{2-} is the conjugate base of \text{H}_2\text{AsO}_4^{-}, its conjugate acid.
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C. Their components can be separated by physical processes.

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Before tackling this problem, be sure you know how to find the antilog of a number using a scientific calculator.
dybincka [34]
<h2>Question:- </h2>

A solution has a pH of 5.4, the determination of [H+].

<h2>Given :- </h2>
  1. pH:- 5.4
  2. pH = - log[H+]

<h2>To find :- concentration of H+</h2>

<h2>Answer:- Antilog(-5.4) or 4× 10-⁶</h2>

<h2>Explanation:- </h2><h3>Formula:- pH = -log H+ </h3>

Take negative to other side

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multiple Antilog on both side

(Antilog and log cancel each other )

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New Formula :- Antilog (-pH) = [+H]

Now put the values of pH in new formula

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we can write -5.4 as (-6+0.6) just to solve Antilog

Antilog ( -6+0.6 ) = [+H]

Antilog (-6) × Antilog (0.6) = [+H]

Antilog (-6)  = {10}^{ - 6} ,  \\ Antilog (0.6)  = 4

put the value in equation

{10}^{ - 6}   \times 4 = [H+] \\ 4 \times   {10}^{ - 6}  = [H+]

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