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iris [78.8K]
2 years ago
9

How many Liters of H2O are in 1.50 g H2O?

Chemistry
2 answers:
zlopas [31]2 years ago
6 0

Answer:18.02 grams

The average mass of one mole of H2O is 18.02 grams.

Explanation: we need to find the moles of water in 1 L of water or 1000mL of water. Taking the density of water to be 1g/mL, 1000mL of water= 1000g water.

Therefore, number of moles= given mass of water(1000)/molecular mass of water(18)

Moles =1000/18= 55.55

So total molecules in 55.55 moles= 55.55* 6.022*10^23 = 334.52*10^23 molecules

=3.34*10^25 molecules of water.

:)

Trava [24]2 years ago
6 0

Answer:

18.02

Explanation:

is your answer

<h2><u>PLZ mark me as a Brainliest</u></h2>
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strojnjashka [21]

Answer:

The correct answer is...

A) 1/2 hour

Explanation:

5 0
3 years ago
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Tetrahydrofuran (THF) is a common organic solvent with a boiling point of 339 K. Calculate the total energy (q) required to conv
poizon [28]

Explanation:

For the given reaction, the temperature of liquid will rise from 298 K to 339 K. Hence, heat energy required will be calculated as follows.

             Q_{1} = mC_{1} \Delta T_{1}

Putting the given values into the above equation as follows.

           Q_{1} = mC_{1} \Delta T_{1}

                      = 27.3 g \times 1.70 J/g K \times 41

                      = 1902.81 J

Now, conversion of liquid to vapor at the boiling point (339 K) is calculated as follows.

           Q_{2} = energy required = mL_{v}

    L_{v} = latent heat of vaporization

Therefor, calculate the value of energy required as follows.

             Q_{2} = mL_{v}

                         = 27.3 \times 444

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Therefore, rise in temperature of vapor from 339 K to 373 K is calculated as follows.

            Q_{3} = mC_{2} \Delta T_{2}

Value of C_{2} = 1.06 J/g,    \Delta T_{2} = (373 -339) K = 34 K

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             Q_{3} = mC_{2} \Delta T_{2}

                       = 27.3 g \times 1.06 J/g \times 34 K

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            Q = Q_{1} + Q_{2} + Q_{3}

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Thus, we can conclude that total energy (q) required to convert 27.3 g of THF at 298 K to a vapor at 373 K is 15007.902 J.

5 0
3 years ago
What is the energy of a photon having a wavelength of 2.60 ✕ 10-8 m?<br> J
zvonat [6]

The energy of a photon : 7.645 x 10⁻³⁴ J

<h3>Further explanation </h3>

Radiation energy is absorbed by photons

The energy in one photon can be formulated as

\large{\boxed{\bold{E\:=\:h\:.\:f}}}

Where

h = Planck's constant (6,626.10⁻³⁴ Js)

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3 0
3 years ago
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Explanation:

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