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lukranit [14]
2 years ago
5

Before tackling this problem, be sure you know how to find the antilog of a number using a scientific calculator.

Chemistry
2 answers:
Marina86 [1]2 years ago
8 0

Answer:

hello

Explanation:

points

dybincka [34]2 years ago
7 0
<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

-pH = log H+

multiple Antilog on both side

(Antilog and log cancel each other )

Antilog (-pH) = [ H+ ]

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

Now put the values of pH in new formula

Antilog (-5.4) = [+H]

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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An 84-mg sample of a compound is found to contain 36 mg of carbon, 3 mg of hydrogen, 21 mg of nitrogen, and 24 mg of oxygen. If
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Answer:

The molecular formula of this compound is C4H4N2O2

Explanation:

Step 1: Data given

Mass of the compound = 84 mg

The compound contains:

36 mg of Carbon

3 mg of hydrogen

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Molar mass of carbon = 12.01 g/mol

Molar mass of hydrogen = 1.01 g/mol

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Molar mass of oxygen = 16 g/mol

Step 2: Calculate number of moles

Moles = mass / molar mass

Moles of carbon = 0.036 g/ 12.01 g/mol = 0.003 moles

Moles of hydrogen = 0.003 g/ 1.01 g/mol = 0.003 moles

Moles of nitrogen = 0.021 g/ 14 g/mol = 0.0015 moles

Moles of oxygen = 0.024 g/ 16 g/mol = 0.0015 moles

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We divide by the smallest amount of moles

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The empirical formula is C2H2NO

The molecular mass of this empirical formula is 56 g/mol

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We have to multiply the empirical formula by n

n = 112 g/mol / 56g/mol = 2

We have to multiply the empirical formula by 2

Molecular formula = 2*(C2H2NO) = C4H4N2O2

The molecular formula of this compound is C4H4N2O2

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