How many grams of water contain 2.50 x 10^24 atoms of hydrogen?

Chemistry

1 answer:

Mama L [17]2 years ago

4 0

The mass of water that contains 2.5×10²⁴ atoms of Hydrogen is 74.79 g

<h3>Avogadro's hypothesis </h3>

From Avogadro's hypothesis,

6.02×10²³ atoms = 2 g of H

Therefore,

2.5×10²⁴ atoms = (2.5×10²⁴ × 2) / 6.02×10²³

2.5×10²⁴ atoms = 8.31 g of H

<h3>How to determine the mass of water </h3>

1 mole of water H₂O = (2×1) + 16 = 18 g
Mass of H in 1 mole of water = 2 g

2 g of H is present in 18 g of water.

Therefore,

8.31 g of H will be present in = (8.31 × 18) / 2 = 74.79 g of water.

Thus, 2.5×10²⁴ atoms of Hydrogen is present in 74.79 g of water.

Learn more about Avogadro's number:

brainly.com/question/26141731

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Arrange the following H atom electron transitions in order of decreasing wavelength of the photon absorbed or emitted:

Katyanochek1 [597]

The decreasing order of wavelengths of the photons emitted or absorbed by the H atom is : b → c → a → d

Rydberg's formula :

$\frac{1}{\lambda} = R_h (\frac{1}{n_1^2}-\frac{1}{n_2^2} )$ ,

where λ is the wavelength of the photon emitted or absorbed from an H atom electron transition from $n_1$ to $n_2$ and $R_h$ = 109677 is the Rydberg Constant. Here $n_1$ and $n_2$ represents the transitions.

(a) $n_1$ =2 to $n_2$ = infinity

$\frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{\infty^2} )$ = 109677/4 [since 1/infinity = 0] Therefore, $\lambda$ = 4 / 109677 = 0.00003647 m