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2 years ago

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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It takes 495.0 kJ of energy to remove 1 mole of electron from an atom on the surface of sodium metal. How much energy does it ta

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Answer:

$\lambda=241.9\ nm$

Explanation:

The work function of the sodium= 495.0 kJ/mol

It means that

1 mole of electrons can be removed by applying of 495.0 kJ of energy.

Also,

1 mole = $6.023\times 10^{23}\ electrons$

So,

$6.023\times 10^{23}$ electrons can be removed by applying of 495.0 kJ of energy.

1 electron can be removed by applying of $\frac {495.0}{6.023\times 10^{23}}\ kJ$ of energy.

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Also,

1 kJ = 1000 J

So,

Energy required = $82.18\times 10^{-20}\ J$

Also, $E=\frac {h\times c}{\lambda}$

Where,

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

So,

$79.78\times 10^{-20}=\frac {6.626\times 10^{-34}\times 3\times 10^8}{\lambda}$

$\lambda=\frac{6.626\times 10^{-34}\times 3\times 10^8}{82.18\times 10^{-20}}$

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Answer:

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In order to arrive at a molecular formula we have to make some assumptions and they are

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next : assume the presence of 6 carbon atoms in said molecule

Total mass = 6 * 12 = 72

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