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4 years ago

Which values do scientists compare when using potassium argon dating

Chemistry

2 answers:

bixtya [17]4 years ago

4 0

The amount of parent and daughter material trapped in the rock.

antoniya [11.8K]4 years ago

4 0

Answer:

Explanation:

edge

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The values used in the scale of pH and pOH are derived from a system designed by ______. Gordonsen Sorenson Curie Dalton

yaroslaw [1]

Sorenson

Explanation:

The values used in the scale of pH and pOH are derived from a system designed by Sorenson. Søren Peter Lauritz Sørensen, a Danish chemist introduced the system of pH and pOH for describing the alkalinity and acidity of substances.

The pH and pOH scale is logarithmic scale that ranks the acidity and bascity of compounds.
pH is the negative logarithm of the concentration of hydrogen/hydroxonium ions in solution i.e

pH = -log₁₀{H⁺]

pOH is the negative log of the concentration of the hydroxyl ions in a solution i.e

pOH = -log₁₀{OH⁻]

Learn more:

calculating pH: brainly.com/question/12985875

pH scale: brainly.com/question/11063271

#learnwithBrainly

3 0

4 years ago

Silver jewelry is a mixture of silver and copper. If a bracelet has a mass of 23.56 g, and it is 80.1% silver, the mass of silve

Reptile [31]

Answer:

18.87 g

Explanation:

80.1% of 23.56 g = 18.87 g is the mass of the silver

6 0

3 years ago

A river with a flow of 50 m3/s discharges into a lake with a volume of 15,000,000 m3. The river has a background pollutant conce

spin [16.1K]

Explanation:

The given data is as follows.

Volume of lake = $15 \times 10^{6} m^{3}$ = $15 \times 10^{6} m^{3} \times \frac{10^{3} liter}{1 m^{3}}$

Concentration of lake = 5.6 mg/l

Total amount of pollutant present in lake = $5.6 \times 15 \times 10^{9} mg$

= $84 \times 10^{9}$ mg

= $84 \times 10^{3}$ kg

Flow rate of river is 50 $m^{3} sec^{-1}$

Volume of water in 1 day = $50 \times 10^{3} \times 86400 liter$

= $432 \times 10^{7}$ liter

Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are $2.9792 \times 10^{10} mg$ or $2.9792 \times 10^{4} kg$

Flow rate of sewage = $0.7 m^{3} sec^{-1}$

Volume of sewage water in 1 day = $6048 \times 10^{4}$ liter

Concentration of sewage = 300 mg/L

Total amount of pollutants = $1.8144 \times 10^{10} mg$ or $1.8144 \times 10^{4}kg$

Therefore, total concentration of lake after 1 day = $\frac{131936 \times 10^{6}}{1.938 \times 10^{10}}mg/ l$

= 6.8078 mg/l

$k_{D}$ = 0.2 per day

$L_{o}$ = 6.8078

Hence, $L_{liquid}$ = $L_{o}(1 - e^{-k_{D}t}$

$L_{liquid}$ = $6.8078 (1 - e^{-0.2 \times 1})$

= 1.234 mg/l

Hence, the remaining concentration = (6.8078 - 1.234) mg/l

= 5.6 mg/l

Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.

5 0

4 years ago

2+5*6<br> how to get over someone<br> me and him were together for 10 months

NeX [460]

You should should on yourself, treat yourself

Hangout with your friends and family, distract yourself

It’s okay to be sad for a couple days but be strong

You got this! :)

8 0

3 years ago

Which of the following is a simple definition of reduction?

marshall27 [118]

Answer:

<h3>the gain of electrons is called reduction..</h3>

Explanation:

<h2>彡jessbragoli</h2>

7 0

3 years ago