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Irina18 [472]
3 years ago
8

Find the mass of 4.55 x 1028 atoms of vanadium (V)

Chemistry
1 answer:
daser333 [38]3 years ago
3 0

Answer:

......................

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If there are 1.55 x 1024 molecules of hydrogen peroxide (H2O2), what is the mass of the<br>sample? ​
bezimeni [28]

Answer:

87.54 g of H₂O₂

Explanation:

From the question given above, the following data were obtained:

Number of molecules = 1.55×10²⁴ molecules

Mass of H₂O₂ =.?

From Avogadro's hypothesis,

6.02×10²³ molecules = 1 mole of H₂O₂

Next, we shall determine the mass of 1 mole of H₂O₂. This can be obtained as follow:

1 mole of H₂O₂ = (2×1) + (2×16)

= 2 + 32

= 34 g

Thus,

6.02×10²³ molecules = 34 g of H₂O₂

Finally, we shall determine mass of H₂O₂ that contains 1.55×10²⁴ molecules. This can be obtained as follow:

6.02×10²³ molecules = 34 g of H₂O₂

Therefore,

1.55×10²⁴ molecules

= (1.55×10²⁴ × 34)/6.02×10²³

1.55×10²⁴ molecules = 87.54 g of H₂O₂

Thus, 87.54 g of H₂O₂ contains 1.55×10²⁴ molecules.

8 0
3 years ago
About 25% of the world's population lives:
Pachacha [2.7K]
The answer is without access to energy
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NOT A SERIOUS QUESTION IM JUST EXPLORING BRAINLY.COM!!
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Well it matters If it touch with liquid or soild, if touched with liquid it is not call wet, but with soild yes because something dry will get went when touched with water but something wet will stay wet when touched with water.

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Each element in the periodic table is assigned an atomic number. This number is the same as ___.
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answer is D

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5000kg of ammonium nitrate per square kilometer of cornfield per year. how much nitric acid would be needed to make the fertiliz
Serggg [28]

NH_3+HNO_3-> NH_4NO_3

1 mole of nitric acid produce 1 mole of ammonium nitrate.

moles in 5000 kg of ammonium nitrate :

n=\dfrac{5000\times 1000}{80}\\\\n=62500\ moles ( molecular mass of ammonium nitrate is 80 gm/mol )

So, number of moles of nitric acid required are also 62500 moles.

Mass of 62500 moles of nitric acid :

mass = 62500\times 63 \\\\mass = 3937500\ gram\\\\mass = 393.75\ kg

Hence, this is the required solution.

5 0
3 years ago
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