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ivolga24 [154]
3 years ago
7

A 2.1−mL volume of seawater contains about 4.0 × 10−10 g of gold. The total volume of ocean water is about 1.5 × 1021 L. Calcula

te the total worth of all the gold in the world's oceans if the price of gold is $20.44 per gram.
Chemistry
1 answer:
Fiesta28 [93]3 years ago
8 0

Answer:

Total worth of gold in the ocean = $5,840,000,000,000,000

Explanation:

As stated in the question above, 4.0 x 10^-10 g of gold was present in 2.1mL of ocean water.

Therefore, In 1 L of ocean water there will be,

(4.0 x 10^-10)/0.0021

= 1.9045 x 10^-7 g of gold per Liter of ocean water.

So in 1.5 x 10^-21 L of ocean water, there will be

(1.9045 x 10^-7) * (1.5 x 10^-21)

= 2.857 x 10^14 g of gold in the ocean.

1 gram of gold costs $20.44, that is 20.44 dollars/gram. The total cost of the gold present in the ocean is

20.44 * (2.857 x 10^14)

= $5,840,000,000,000,000

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B/A hope it help thanks
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3 years ago
Q2.  0.254 g of KHP (204 g/mol) titrated against 20.0 mL of unknown NaOH (40.0 g/mol) solution to get the end point of phenolpht
Vera_Pavlovna [14]

Answer:

<u>Mass concentration (g/L) </u><u><em>= 2.49g/L.</em></u>

Explanation:

No. of moles = \frac{mass}{molar mass}

= \frac{0.254}{204} = 0.001245 moles

Concentration of KHP (C1) in litres = n/v

= \frac{0.001245}{0.02} = 0.062 mol/L

We know that:

C_{1} V_{1} = C_{2} V_{2}

where c1v1 and c2v2 are the products of concentration and volumes of KHP and NaOH respectively.

Since mole ratio is 1 : 1.

1 mole of NaOH - 40g

0.001245 mole of NaOH = 40 × 0.001245 = 0.0498g

⇒0.0498g of NaOH was used during the titration

<u><em>∴Mass concentration (g/L) = 0.0498g ÷ 0.02L</em></u>

<u><em>= 2.49g/L.</em></u>

3 0
3 years ago
What’s the difference between sucrose, fructose and glucose?
Sergeeva-Olga [200]

Answer:

Explanation:

Sucrose Is Made up of Glucose and Fructose

Glucose is a simple sugar or monosaccharide.

Fructose, or “fruit sugar,” is a monosaccharide like glucose

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3 years ago
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suter [353]

Answer:

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Explanation:

Hope i helped~  

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3 years ago
Write a net ionic equation for the reaction that occurs when excess hydrochloric acid (aq) and potassium sulfite (aq) are combin
Ket [755]

<u>Answer:</u> The net ionic equation for the given reaction is 2H^+(aq.)+SO_3^{2-}(aq.)\rightarrow H_2O(l)+SO_2(g)

<u>Explanation:</u>

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are the ions which do not get involved in a chemical equation. It is also defined as the ions that are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of hydrochloric acid and potassium sulfite is given as:

2HCl(aq.)+K_2SO_3(aq.)\rightarrow 2KCl(aq.)+SO_2(g)+H_2O(l)

Ionic form of the above equation follows:

2H^+(aq.)+2Cl^-(aq.)+2K^+(aq.)+SO_3^{2-}(aq.)\rightarrow 2K^+(aq.)+2Cl^-(aq.)+SO_2(g)+H_2O(l)

As, potassium and chloride ions are present on both the sides of the reaction, thus, it will not be present in the net ionic equation.

The net ionic equation for the above reaction follows:

2H^+(aq.)+SO_3^{2-}(aq.)\rightarrow SO_2(g)+H_2O(l)

Hence, the net ionic equation for the given reaction is written above.

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