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3 years ago

Select the correct answer.

Chemistry

1 answer:

ankoles [38]3 years ago

4 0

It would still have oceans but no atmospheric water in Earth if no icy debris had arrived.

A. It would still have oceans but no atmospheric water.

<u>Explanation:</u>

Seas characterize our home planet, covering most of the Earth's surface and driving the water cycle that commands our territory and climate. However, progressively significant still, the narrative of our seas wraps our home in a far bigger setting that ventures profound into the universe and spots us in a rich group of sea universes that range our nearby planetary group and past.

It would in any case have seas yet no air water on Earth if no frigid flotsam and jetsam had shown up. For a long time, it was accepted that the frosty moons were only that - solidified husks, strong to their center. However, lately that thought has steadily been supplanted by a fresher, additionally energizing worldview.

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A student followed the procedure of this experiment to determine the solubility product of zinc(II) iodate, Zn(IO3)2. Solutions

Fed [463]

Answer:

Explanation:

Complete the table below and determine the solubility product constant.

[Zn(NO₃)₂] 0, M Initial 0.226, 0.101 0.0452, 0.0118

[KIO₃] 0, M Titrant 0.200, 0.200, 0.200, 0.200

V₀, mL of Zn(NO₃)₂100.0, 100.0, 100.0, 100.0

V, mL of KIO₃ titrant 12.9, 12.4, 13.0, 18.3

What needs to be determined are the following for each column/sample

V0 + V, mL

[Zn²⁺] + [IO³⁻] ⇄ [Zn²⁺][IO³⁻]2

log [Zn²⁺][IO³⁻]2

$\frac{\sqrt{Zn^{2+}}}{1 + \sqrt{Zn^{2+}}}$

Then there's the determination of Ksp itself

log Ksp = Ksp of Zn(IO3)2 =

Considering this, which of the ion products is closest to Ksp, and why?

Finally, the titration volumes for the first three samples don't vary greatly. However the last sample is considerably larger. Why is this to be expected?

The attached figures shed more light on the solution to this problem

7 0

3 years ago

137g of Ba + ______ g oh I --> 391g of BaI2

sertanlavr [38]

Answer:

Explanation:

Is it worth it? Joy

7 0

3 years ago

The veterinarian prescribes the NSAID carprofen for a 50-pound dog with arthritis at a dose of 4.4 mg/kg and would like this dos

alina1380 [7]

Answer:

The dog should be given half of 100 mg tablet in the morning and another half of 100 mg tablet in an evening.

Explanation:

Weight of the dog = 50 pounds = 22.68 kg

1 kg = 2.205 pounds

Amount of daily dose prescribed by doctor = 4.4 mg/kg

Total mount of dose = 4.4 mg/kg × 22.68 kg = 99.79 mg

99.79 mg ≈ 100 mg

49.89 mg ≈ 50 mg

So, according to doctor prescription dog should be given half 100 mg tablet in the morning and another half of 100 mg tablet in an evening.

8 0

3 years ago

Whag is the answer. please i need help

icang [17]

Answer:

First option

Explanation:

Students 1 and 2 have precise measures yet is is not accurate since they measure around 5.

5 0

3 years ago

Ammonium phosphate nh43po4 is an important ingredient in many fertilizers. it can be made by reacting phosphoric acid h3po4 with

Tom [10]

The reaction between phosphoric acid and ammonia that produces ammonium phosphate can be written as follows:
3NH3 + H3PO4 ..................> (NH4)3PO4

From the periodic table:
molar mass of nitrogen = 14 grams
molar mass of hydrogen = 1 grams
molar mass of oxygen = 16 grams
molar mass of phosphorus = 30.9 grams

based on this:
molar mass of 3NH3 = 3 (14 + 3(1)) = 51 grams
molar mass of H3PO4 = 3(1) + 30.9 + 4(16) = 97.9 grams
molar mass of (NH4)3PO4 = 3 (14 + 4(1)) + 30.9 + 4(16) = 54 + 30.9 + 64
= 148.9 grams

Therefore, 97.9 grams of phosphoric acid is required to produced 148.9 grams of ammonium phosphate.
Thus, to know the mass of ammonium phosphate produced from 4.9 grams of phosphoric acid, we will simply use cross multiplication as follows:
amount of produced ammonium phosphate = (4.9 x 148.9) / 97.9 = 7.45 g

3 0

3 years ago

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