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4 years ago

5

What is absolute dating?

2 answers:

KiRa [710]4 years ago

4 0

Absolute dating is finding the exact date of an artifact such as a fossil using a technique such as radioactive dating, etc.

Varvara68 [4.7K]4 years ago

4 0

Explanation:

Absolute dating is the process of determining an age on a specified chronology in archaeology and geology. Some scientists prefer the terms chronometric or calendar dating, as use of the word "absolute" implies an unwarranted certainty of accuracy.

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I will give brainliest and 76 points

katen-ka-za [31]

Answer:

number 5 is the answer

Explanation:

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3 years ago

(C) A metal 'M' has electronic configuration 2, 8, 2. Find the formula of its

borishaifa [10]

M = 2 . 8 . 2

Valence Electron of M = 2

M ==> M⁺² + 2 e⁻

a. M⁺² + OH⁻ ==> M(OH)₂

b. M⁺² + PO₄⁻³ ==> M₃(PO₄)₂

7 0

3 years ago

The upward force that acts on a birds wings due to its flight is called

Radda [10]

The answer is lift. (d)

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3 years ago

If a 95.27 mL sample of acetic acid (HC2H3O2) is titrated to the equivalence point with 79.06 mL of 0.113 M KOH, what is the pH

KATRIN_1 [288]

Answer:

8.73

Explanation:

The concentration of acetic acid can be determined as follows:

$M_1V_1 = M_2V_2\\(KOH) = (CH_3COOH)$

$M_{KOH}=0.113 M\\V_{KOH}=79.06 mL$

$V_{CH_3COOH}=95.27 \\\\M_{CH_3COOH)=?????$

$M_{CH3COOH} = \frac{M_{KOH}*V_{KOH}}{V_{CH_3COOH}}$

$M_{CH3COOH} = \frac{0.113*79.06}{95.27}$

$M_{CH3COOH} = 0.094 M$

Moles of $CH_3COOH$ = $95.27* 10^{-3}* 0.094$

=0.0090 moles

Moles of $KOH = 79.06*10^{-3}*0.113$

= 0.0090 moles

The equation for the reaction can be expressed as :

$CH_3COOH$ $+$ $KOH$ -----> $CH_3COO^{-}K^+$ $+$ $H_2O$

Concentration of $CH_3COO^{-$ ion = $\frac{0.0090}{Total volume (L)}$

= $\frac{0.0090}{(95.27+79.06)} *1000$

= 0.052 M

Hydrolysis of $CH_3COO^{-$ ion:

$CH_3COO^{-$ $+$ $H_2O$ -----> $CH_3COOH$ $+$ $OH^-$

$K = \frac{K_w}{K_a} = \frac{x*x}{0.052-x}$

⇒ $\frac{10^{-14}}{1.82*10^{-5}}= \frac{x*x}{0.052-x}$

= $0.5494*10^{-9}= \frac{x*x}{0.052-x}$

As K is so less, then x appears to be a very infinitesimal small number

0.052-x ≅ x

$0.5494*10^{-9}= \frac{x^2}{0.052}$

$x^2 = 0.5494*10^{-9}*0.052$

$x^2 = 0.286*10^{-10$

$x = \sqrt{0.286*10^{-10$

$x =0.535*10^{-5}M$

$[OH] = x =0.535*10^{-5}$

$pOH = -log[OH^-]$

$pOH = -log[0.535*10^{-5}]$

$pOH = 5.27$

pH = 14 - pOH

pH = 14 - 5.27

pH = 8.73

Hence, the pH of the titration mixture = 8.73

8 0

3 years ago

Match the substance with its chemical formula.

ch4aika [34]

H 3 0 + is hydronium ion. OH- is hydroxide ion. H+ is hydrogen ion

4 0

4 years ago

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