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3 years ago

What evidence exists to support the idea that the composition of ocean water includes dissolved oxygen?

Chemistry

2 answers:

blondinia [14]3 years ago

6 0

Answer:

The evidence that exists to support the idea that the composition of ocean water includes dissolved oxygen is that all water has oxygen in it.

Explanation:

Setler79 [48]3 years ago

5 0

Answer:

Fish get oxygen from water through their gills - they require oxygen to survive, so they have to be getting it from the water.

Explanation:

The concentration of dissolved oxygen and carbon dioxide are very important for marine life forms. Although both oxygen and carbon dioxide are a gas when outside the water, they dissolve to a certain extent in liquid seawater. Dissolved oxygen is what animals with gills use for respiration (their gills extract the dissolved oxygen from the water flowing over the gill filaments).

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Suppose of ammonium nitrate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of ammonium c

Marta_Voda [28]

Answer:

Final molarity of ammonium cation in the solution = 0.16 M

Explanation:

Complete Question

Suppose 2.59 g of ammonium nitrate is dissolved in 200. mL of a 0.40M aqueous solution of sodium chromate. Calculate the final molarity of ammonium cation in the solution. You can assume the volume of the solution doesn't change when the ammonium nitrate is dissolved in it. Be sure your answer has the correct number of significant digits.

Solution

2NH₄NO₃ + Na₂CrO₄ → (NH₄)₂CrO₄ + 2NaNO₃

We first convert the given parameters to number of moles

Number of moles = (Mass/Molar mass)

Molar mass of NH₄NO₃ = 80.043 g/mol

Number of moles of NH₄NO₃ = (2.59/80.043) = 0.03224 mole

Number of moles = (Concentration in mol/L) × (Volume in L)

Number of moles of Na₂CrO₄ = 0.4 × 0.2 = 0.08 Mole

2 moles of NH₄NO₃ react with 1 mole of Na₂CrO₄

So, it it evident that NH₄NO₃ is the limiting reagent as it is in short supply in the amount needed for the reaction.

So, the number of moles of ammonium ion in the product is also 0.03224 mole.

Molarity = (Number of moles)/(Volume L)

Molarity of ammonium ion = (0.03224/0.2) = 0.1612 mol/L = 0.16 M

Hope this Helps!!!

3 0

2 years ago

Molodets [167]

False. The answer is liter. kilo is the base (liter) times 1000. it would make no sense.

5 0

3 years ago

How many protons does one atom of sulfur have

mrs_skeptik [129]

Answer:

Explanation:

The proton number is always the number in the corner of the box on a periodic table

6 0

3 years ago

Read 2 more answers

How many grams of Sg is required to produce 83.10 g SF6? S: +24F-->8SF

ozzi

Answer : The mass of $S_8$ required is 18.238 grams.

Explanation : Given,

Mass of $SF_6$ = 83.10 g

Molar mass of $SF_6$ = 146 g/mole

Molar mass of $S_8$ = 256.52 g/mole

The balanced chemical reaction is,

$S_8+24F_2\rightarrow 8SF_6$

First we have to determine the moles of $SF_6$ .

$\text{Moles of }SF_6=\frac{\text{Mass of }SF_6}{\text{Molar mass of }SF_6}=\frac{83.10g}{146g/mole}=0.569moles$

Now we have to determine the moles of $S_8$ .

From the balanced chemical reaction we conclude that,

As, 8 moles of $SF_6$ produced from 1 mole of $S_8$

So, 0.569 moles of $SF_6$ produced from $\frac{0.569}{8}=0.0711$ mole of $S_8$

Now we have to determine the mass of $S_8$ .

$\text{Mass of }S_8=\text{Moles of }S_8\times \text{Molar mass of }S_8$

$\text{Mass of }S_8=(0.0711mole)\times (256.52g/mole)=18.238g$

Therefore, the mass of $S_8$ required is 18.238 grams.

7 0

3 years ago

Why did bohrs observations of gaseous elements not reveal a complete spectrum when the light was shone through a prism? explain

Vika [28.1K]

The Bohr model proposed that electrons could just have characterized vitality levels thus when rotting back to a lower vitality level discharge a specific measure of vitality. Since the measure of vitality could be changed over to a specific recurrence then particular emanation lines were found in the electromagnetic range. Alternate speculations couldn't clarify the discharge lines.

4 0

3 years ago