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3 years ago

PLZ HELP

Chemistry

2 answers:

nika2105 [10]3 years ago

8 0

Mjdksksksnsjdkjdjsksksjdndnd

denis23 [38]3 years ago

3 0

I do not believe you can tell the displacement by the information given. so i would say none of the above.

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How do you toast a toaster?

Mama L [17]

by putting to much current through it ?

7 0

3 years ago

Pickles are cucumbers preserved in a brine solution containing dill, alum, and salt. Cucumbers shrink to a fraction of their nor

Anna [14]

Answer:

hypertonic solution

Explanation:

Hypertonic solution -

It is the solution, with more amount of solute than the solvent , is known as hypertonic solution.

Now, is some substance is immersed in such solution , the substance gets shrinked , because , the solvent from the substance moves out of it and moves to the hypertonic solution.

Hence, the pickles gets shrinked up , as they put in a hypertonic solution.

8 0

3 years ago

Iridium has two naturally occurring isotopes, Iridium-191 and Iridium-193. If the average atomic mass of orodum s 192.217, what

Lina20 [59]

Answer:

The percent isotopic abundance of Ir-193 is 60.85 %

The percent isotopic abundance of Ir-191 is 39.15 %

Explanation:

we know there are two naturally occurring isotopes of iridium, Ir-191 and Ir-193

First of all we will set the fraction for both isotopes

X for the isotopes having mass 193

1-x for isotopes having mass 191

The average atomic mass is 192.217

we will use the following equation,

193x + 191(1-x) = 192.217

193x + 191 - 191x = 192.217

193x- 191x = 192.217 - 191

2x = 1.217

x= 1.217/2

x= 0.6085

0.6085 × 100 = 60.85 %

60.85% is abundance of Ir-193 because we solve the fraction x.

now we will calculate the abundance of Ir-191.

(1-x)

1-0.6085 =0.3915

0.3915× 100= 39.15 %

6 0

3 years ago

Calculate the volume of the object immersed in the water in the water in the adjoining figur.

kirza4 [7]

Answer:

100 cm³

Explanation:

Hi there!

Subtract:

400 cm³ - 300 cm³

= 100 cm³

Therefore, the volume of the object is 100 cm³.

I hope this helps!

5 0

3 years ago

What mass of precipitate forms when 185.5 ml of 0.533 m naoh is added to 627 ml of a solution that contains 15.8 g of aluminum s

Law Incorporation [45]

Answer:

2,57 g of precipitate.

Explanation:

For the reaction:

6 NaOH + Al₂(SO₄)₃ → 2 Al(OH)₃ + 3 Na₂SO₄

The precipitate is Al(OH)₃.

185,5mL of 0,533M NaOH are:

0,1855L × 0,533M = 0,0989 moles NaOH

Moles of Al₂(SO₄)₃ are:

15,8g × $\frac{1mol}{342,15g}$ = 0,0462 moles Al₂(SO₄)₃

For the total reaction of 0,0989 moles NaOH with Al₂(SO₄)₃ you need:

0,0989moles NaOH × $\frac{1molAl_{2}(SO_{4})_{3}}{6 moles NaOH}$ = 0,0165 moles Al₂(SO₄)₃

As you have 0,0462 moles Al₂(SO₄)₃ the limiting reactant is NaOH.

0,0989 moles of NaOH produce:

0,0989moles NaOH × $\frac{2molAl(OH)_{3}}{6 moles NaOH}$ = 0,0330 moles of Al(OH)₃

These moles are:

0,0330 moles of Al(OH)₃ × (78 g/mol) = 2,57 g of Al(OH)₃ ≡ mass of precipitate

I hope it helps!

3 0

3 years ago