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umka2103 [35]
3 years ago
11

[H3O+] [OH−] pH Acidic or Basic 3.5×10−3 _____ _____ _____ _____ 3.8×10−7 _____ _____ 1.8×10−9 _____ _____ _____ _____ _____ 7.1

5 _____[H3O+] [OH−] pH Acidic or Basic 3.5×10−3 _____ _____ _____ _____ 3.8×10−7 _____ _____ 1.8×10−9 _____ _____ _____ _____ _____ 7.15 _____

Chemistry
1 answer:
ankoles [38]3 years ago
7 0

Answer:

See explanation below

Explanation:

I'm assuming this is a table you need to complete, so, you are not putting this in order, but I already found it in another place. let's do a little summary of the expressions we need to use in order to complete the chart.

To calculate pH we need the following expression:

<em>pH = -log[H3O+] (1)</em>

From this expression we can solve for [H3O+] in case we need it:

<em>[H3O+] = 10^(-pH)   (2)</em>

When we have [OH-] we calculate the pOH and from there, the pH:

<em>pOH = -log[OH-]  (3)</em>

and [OH-]:

<em>[OH-] = 10^(-pOH) (4)</em>

Finally to get the pH from pOH:

14 = pH + pOH

<em>pH = 14 - pOH (5)</em>

With these 5 expressions we can complete the chart. In picture 1, you have the actual chart.

To know if it's acidic or basic, that depends on the value of pH.

If pH <7 it's acidic

If pH >7 it's basic

If pH = 7 it's neutral

<u><em>First case:</em></u>

[H3O+] = 3.5x10^-3

pH = -log(3.5x10^-3) = 2.46  (acidic)

pOH = 14 - 2.46 = 11.54

[OH-] = 10^(-11.54) = 2.88x10^-12 M

<u><em>Second case:</em></u>

pOH = -log(3.8x10^-7) = 6.42

pH = 14 - 6.42 = 7.58 (it's basic)

[H3O+] = 10^(-7.58) = 2.63x10^-8 M

<u><em>Third case:</em></u>

pH = -log(1.8x10^-9) = 8.74 (it's basic)

pOH = 14 - 8.74 = 5.26

[OH-] = 10^(-5.26) = 5.5x10^-6 M

<u><em>Fourth case:</em></u>

[H3O+] = 10^(-7.15) =7.08x10^-8 M   (Basic)

pOH = 14 - 7.15 = 6.85

[OH-] = 10^(-6.85) = 1.41x10^-7 M

Hope this can help you

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A.) A student titrated a 15.00-mL sample of a solution containing a weak, monoprotic acid with NaOH. If the titration required 1
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Answer:

A) 0.1225 M

B) 100.4 g/mol

Explanation:

Step 1: Write the generic neutralization reaction

HA(aq) + NaOH(aq) ⇒ NaA(aq) + H₂O(l)

Step 2: Calculate the reacting moles of NaOH

17.73 mL of 0.1036 M NaOH react. The reacting moles are:

0.01773 L × 0.1036 mol/L = 1.837 × 10⁻³ mol

Step 3: Calculate the reacting moles of HA

The molar ratio of HA to NaOH is 1:1. The reacting moles of HA are 1/1 × 1.837 × 10⁻³ mol = 1.837 × 10⁻³ mol.

Step 4: Calculate the molar concentration of HA

1.837 × 10⁻³ moles of HA are in a 15.00 mL volume. The molar concentration is:

M = 1.837 × 10⁻³ mol / 0.01500 L = 0.1225 M

Step 5: Calculate the molar mass of HA

1.837 × 10⁻³ moles of HA weigh 0.1845 g. The molar mass of HA is:

0.1845 g / 1.837 × 10⁻³ mol = 100.4 g/mol

5 0
3 years ago
Based on the molecular formula, determine whether each of the following is an alkane, alkene, or alkyne. (Assume that the hydroc
stealth61 [152]

Answer:

C₃H₆ is an alkene

C₆H₁₂ is an alkene

C₈H₁₈ is an alkane

C₇H₁₂ is an alkyne

Explanation:

To determine which of the compound is alkane, alkene, or alkyne,we shall use the general formula of alkane, alkene, and alkyne. This is illustrated below:

General formula for alkane => CₙH₂ₙ₊₂

General formula for alkene => CₙH₂ₙ

General formula for alkyne => CₙH₂ₙ₋₂

For C₃H₆:

n = 3

Alkane => CₙH₂ₙ₊₂ => C₃H₂₍₃₎₊₂ => C₃H₈

Alkene => CₙH₂ₙ => C₃H₂₍₃₎ => C₃H₆

Alkyne => CₙH₂ₙ₋₂ => C₃H₂₍₃₎₋₂ => C₃H₄

Thus, C₃H₆ is an alkene

For C₆H₁₂:

n = 6

Alkane => CₙH₂ₙ₊₂ => C₆H₂₍₆₎₊₂ =>C₆H₁₄

Alkene => CₙH₂ₙ => C₆H₂₍₆₎ => C₆H₁₂

Alkyne => CₙH₂ₙ₋₂ => C₆H₂₍₆₎₋₂ => C₆H₁₀

Thus, C₆H₁₂ is an alkene

For C₈H₁₈:

n = 8

Alkane => CₙH₂ₙ₊₂ => C₈H₂₍₈₎₊₂ => C₈H₁₈

Alkene => CₙH₂ₙ => C₈H₂₍₈₎ => C₈H₁₆

Alkyne => CₙH₂ₙ₋₂ => C₈H₂₍₈₎₋₂ => C₈H₁₄

Thus, C₈H₁₈ is an alkane.

For C₇H₁₂:

n = 7

Alkane => CₙH₂ₙ₊₂ => C₇H₂₍₇₎₊₂ => C₇H₁₆

Alkene => CₙH₂ₙ => C₇H₂₍₇₎ => C₇H₁₄

Alkyne => CₙH₂ₙ₋₂ => C₇H₂₍₇₎₋₂ => C₇H₁₂

Thus, C₇H₁₂ is an alkyne.

SUMMARY:

C₃H₆ is an alkene

C₆H₁₂ is an alkene

C₈H₁₈ is an alkane

C₇H₁₂ is an alkyne

3 0
3 years ago
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