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2 years ago

15

Which of the following is the closest to the observed abundance of 71Ga, one of two stable gallium isotopes (69Ga and 71Ga). The

answer in the book shows 40%, and I have absolutely no idea how to solve this. Please help!

1 answer:

natali 33 [55]2 years ago

3 0

The outcome of that will be 44

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What is the pH of solution if the concentration of [H3O+] = 0.000558?

lina2011 [118]

Answer:

3.25

Explanation:

pH = -log[H3O+]

-log(0.000558)=3.25

7 0

3 years ago

The amount of energy available to do work after a chemical reaction has occurred is called

sveticcg [70]

Free energy is the answer i hope this helped

8 0

3 years ago

Sulfur and oxygen form both sulfur dioxide and sulfur trioxide. When samples of these were decomposed the sulfur dioxide produce

Zinaida [17]

Answer : The mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide is, 0.997 g and 1.5 g respectively.

Explanation : Given,

Mass of oxygen in sulfur dioxide = 3.49 g

Mass of sulfur in sulfur dioxide = 3.50 g

Mass of oxygen in sulfur trioxide = 9.00 g

Mass of sulfur in sulfur trioxide = 6.00 g

Now we have to calculate the mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide.

Mass of oxygen per gram of sulfur for sulfur dioxide = $\frac{\text{Mass of oxygen}}{\text{Mass of sulfur}}$

Mass of oxygen per gram of sulfur for sulfur dioxide = $\frac{3.49}{3.50}=0.997g$

and,

Mass of oxygen per gram of sulfur for sulfur trioxide = $\frac{\text{Mass of oxygen}}{\text{Mass of sulfur}}$

Mass of oxygen per gram of sulfur for sulfur trioxide = $\frac{9.00}{6.00}=1.5g$

Thus, the mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide is, 0.997 g and 1.5 g respectively.

8 0

3 years ago

What is the molarity of the bleach solution of slide 28 (7.4% NaOCl by mass, density 1.12 g.cm-3)?

Anuta_ua [19.1K]

Answer:

Molarity of solution is 1.10x10⁻³ M

Explanation:

Solute NaOCl

7.4% by mass means, that in 100 grams of solution, we have 7.4 g of solute.

Molar mass of NaOCl = 74.45 g/m

Mol = Mass / Molar mass

7.4 g / 74.45 g/m = 0.099 moles

Density of solution = 1.12 g/mL

Density = Mass / volume

1.12g/mL = 100 g / volume

Volume = 100 g / 1.12g/mL = 89.3 mL

Molarity = mol /L

89.3 mL = 0.0893 L

0.099 moles / 0.0893 L = 1.10x10⁻³ M

3 0

3 years ago

Read 2 more answers

In which of these cases would it be most useful to use a mole to calculate the number of particles?

Ivan

I believe it would be the last one because you can use the molar mass of HCl to find the number of moles, then use Avogadro’s number to find the number of atoms
Hope this helps!

6 0

3 years ago