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Kryger [21]
2 years ago
15

How many molecules of oxygen are needed to make 3.5 oxygens of water?​

Chemistry
1 answer:
max2010maxim [7]2 years ago
8 0

Answer:

57.6g

Explanation:

So, if in one mole of water, 16 g of oxygen atom is present. Then, in 3.6 moles of water, the mass of oxygen present will be 3.6×16=57.6g. Therefore, the amount of oxygen present in 3.6 g water is option (B)- 57.6 g.

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Is 334mm bigger than 1 km
Levart [38]

Answer:

no km are bigger

Explanation:than mm

5 0
3 years ago
Maple syrup has a density of 1.325 g/ml, and 100.00 g of maple syrup contains 67 mg of calcium in the form of ca2+ ions. what is
e-lub [12.9K]
Density of maple syrup = 1.325 g/ml
1000 ml contains 1325 g of maple syrup
In 100 g of maple syrup - 67 mg of Ca ions
Therefore in 1325 g of maple syrup - 67 mg /100g * 1325 g 
                                                        = 887.75 mg of Ca
this means in 1000 ml - 887.75 mg of Ca
molar mass of Ca - 40 g/mol
therefore number of moles in 1000 ml - 0.88775 g /40 g/mol
molarity of Ca - 0.022 mol/dm³
7 0
4 years ago
Read 2 more answers
A thermally isolated system is made up of a hot piece of aluminum and a cold piece of copper; the aluminum and the copper are in
Illusion [34]

Answer:

  • <u>Copper</u><em> is the object that experiences the greater temperature change.</em>

Explanation:

<em>Thermally isolated system </em>means that the system does not exchange thermal energy with the surroundings.

Hence, any thermal exchange, in virtue of the temperature difference of the aluminum and copper pieces, is between them.

In consequence, the law of conservation of energy states that the heat lost by the hot substance will be gained by the cold matter.

In equations, that is:

  • Heat lost by aluminum = heat gained by copper.

Now, the gain or loss or heat of a substance, Q, is related with the mass (m), the specific heat (Cs), and the cahnge of temperature (ΔT), per the equation:

  • Q = m × Cs × ΔT

∴ Q lost by aluminum = Q  gained copper ⇒

  • [m × Cs × ΔT ] aluminum = [m × Cs × ΔT ] copper.

Under the reasoning assumption that the masses of aluminum and copper are equal, the equations is simplified to:

  • [Cs × ΔT ] aluminum = [Cs × ΔT ] copper.

  • Cs aluminum / Cs copper = ΔT copper / ΔT aluminum

  • Cs aluminum > 2 × Cs copper  ⇒ Cs

  • Cs aluminum / Cs copper > 2 ΔT copper / ΔT aluminum

  • ΔT copper / ΔT aluminum > 2

  • ΔT copper > 2 × ΔT aluminum

In words, since it is stated that  the specific heat of aluminum is more than double that of copper,  in order to keep the equality, ΔT of copper shall be more than double ΔT of aluminum.

Hence, the <u>conclusion</u> is that the object that experiences the greater temperature change is copper (the one with the lower specific heat), under the assumption that both objects have the same amount of matter (mass).

6 0
3 years ago
Provide one example of how life on Earth would not be possible without water based on its characteristics.
snow_lady [41]

Answer: ice is less dense than liquid water. If ice was more dense, Earth would freeze.

Explanation: There are many reasons why life on Earth depends on the characteristics of water. One could discuss hydrogen bonds and its role as a solvent, but the unusual property of water is is the change in density with change in temperature. Water is densest at 4 degC, which is why ice floats - it is less dense than cold water (it melts quickly in warm water, so density isn’t impotant at higher temperatures). Most liquids are less dense than the solid, frozen form. If this was the case with water, any ice that formed would sink, and sease would freeze from the bottom up. Furthermore, the lowest layers would be insulated and would not all melt in summer. Thus over time, the seas would become a thin layer of liquid water at best, over solid ice. Life could not develop without liquid seas. In addition, ice is reflective, reducing the amount of sunlight absorbed, further reducing temperatures. Without ocean circulation, polar areas would be even colder, and there would be no rain.

6 0
3 years ago
You notice that the water in your friend's swimming pool is cloudy and that the pool walls are discolored at the water line. A q
Flauer [41]

Answer:

6,78 mL of 12,0 wt% H₂SO₄

Explanation:

The equilibrium in water is:

H₂O (l) ⇄ H⁺ (aq) + OH⁻ (aq)

The initial concentration of [H⁺] is 10⁻⁸ M and final desired concentration is [H⁺] = 10^{-7,20}, thus,

Thus, you need to add:

[H⁺] = 10^{-7,2} -10^{-8,0} = 5,31x10⁻⁸ M

The total volume of the pool is:

9,00 m × 15,0 m ×2,50 m = 337,5 m³ ≡ 337500 L

Thus, moles of H⁺ you need to add are:

5,31x10⁻⁸ M × 337500 L = 1,792x10⁻² moles of H⁺

These moles comes from

H₂SO₄ → 2H⁺ +SO₄²⁻

Thus:

1,792x10⁻² moles of H⁺ × \frac{1H_{2}SO_4 mol}{2H^{+} mol} = 8,96x10⁻³ moles of H₂SO₄

These moles comes from:

8,96x10⁻³ moles of H₂SO₄ × \frac{98,1 g}{1 mol} × \frac{100 gSolution}{12 gH_{2}SO_4 } × \frac{1 mL}{1,080 g}  =

6,78 mL of 12,0wt% H₂SO₄

I hope it helps!

8 0
3 years ago
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