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3 years ago

What five SI base units are commonly used in chemistry?

Chemistry

2 answers:

Stells [14]3 years ago

8 0

The five most commonly used are:
•Meter
•Kilometer
•Kelvin
•Second
•Mole

Rashid [163]3 years ago

3 0

Five base SI units that are commonly used in Chemistry; are
1.Metre (m)
2. Kilogram(kg)
3. seconds(s)
4.Kelvins (k)
5. Moles (mol)

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Which property of isotopes makes them suitable for determining the dates of ancient artifacts? atomic number of the isotope rate

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Each isotope has a unique rate of decay, making them suitable for determining the dates of ancient artifacts. The answer is "rate of decay of the isotope."

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3 years ago

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3 years ago

A chemist must prepare 800ml of potassium hydroxide solution with a pH 13 of at 25. He will do this in three steps:

Naily [24]

Answer:

4.48 grams of potassium hydroxide that the chemist must be weighing out.

Explanation:

The pH of the KOH solution = 13

pH + pOH = 14

pOH = 14 - pH = 14 - 13 = 1

$pOH=-\log[OH^-]$

$1=-\log[OH^-]$

$[OH^-]=0.1 M$

$KOH(aq)\rightarrow K^+(aq)+OH^-(aq)$

1 mole of hydroxide ions are obtained from 1 mole of KOH. Then 0.1 mole of hydroxide ions will be obtained from :

$\frac{1}{1}\times 0.1 M=0.1 M$ of KOH

$[Molarity]=\frac{\text{Moles of solute}}{\text{Volume of solution(L)}}$

Volume of KOH solution = 800 mL = 0.800 L ( 1 mL = 0.001 L)

$0.1 M=\frac{\text{Moles of KOH}}{0.800 L}$

Moles of KOH = 0.1 M × 0.800 L = 0.08 mol

Mass of 0.08 moles of KOH :

0.08 mol × 56 g/mol = 4.48 g

4.48 grams of potassium hydroxide that the chemist must be weighing out.

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3 years ago

Which of the following are changing when a change of state is occurring?

Greeley [361]

Answer:

What are the options?

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3 years ago

How many grams of a 23.4% by mass naf solution is needed if you want to have 1.33 moles of naf?

jenyasd209 [6]

Percentage by mass is the mass of NaF present in 100 g of the solution.
the percentage by mass of NaF is 23.4 %
this means that in 100 g of solution, mass of NaF present is 23.4 g
the number of moles of NaF present - 1.33 mol
mass of NaF - 1.33 mol x 42 g/mol = 55.9 g
when there's 23.4 g of NaF - mass of solution is 100 g
therefore when there's 55.9 g of NaF - mass of solution is 100 / 23.4 x 55.9
= 239 g
mass of solution required is 239 g

6 0

4 years ago