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Zolol [24]
4 years ago
13

Number of grams of hydrogen than can be prepared from 6.80g of aluminum​

Chemistry
1 answer:
Anastaziya [24]4 years ago
5 0

Answer:

0.7561 g.

Explanation:

  • The hydrogen than can be prepared from Al according to the balanced equation:

<em>2Al + 6HCl → 2AlCl₃ + 3H₂,</em>

It is clear that 2.0 moles of Al react with 6.0 mole of HCl to produce 2.0 moles of AlCl₃ and 3.0 mole of H₂.

  • Firstly, we need to calculate the no. of moles of (6.8 g) of Al:

no. of moles of Al = mass/atomic mass = (6.8 g)/(26.98 g/mol) = 0.252 mol.

<em>Using cross multiplication:</em>

2.0 mol of Al produce → 3.0 mol of H₂, from stichiometry.

0.252 mol of Al need to react → ??? mol of H₂.

∴ the no. of moles of H₂ that can be prepared from 6.80 g of aluminum = (3.0 mol)(0.252 mol)/(2.0 mol) = 0.3781 mol.

  • Now, we can get the mass of H₂ that can be prepared from 6.80 g of aluminum:

mass of H₂ = (no. of moles)(molar mass) = (0.3781 mol)(2.0 g/mol) = 0.7561 g.

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The claim: "If the nucleus were the size of a grape, the electrons would be one mile away on average" is reasonably accurate because the ratios between the nucleus's sizes and the distances (between electrons and nucleus) for the two given examples are in the same order of magnitude.      

To know if the claim is accurate we need to calculate the ratio of the size of the nucleus (the same as a grape) and the distance between the electrons and the nucleus for example 1 (r₁):  

r_{1} = \frac{s_{1}}{d_{1}}    (1)

and to compare it with the ratio of the size and the distance given in example 2 (r₂):

r_{2} = \frac{s_{2}}{d_{2}}    (2)

<em>Where:</em>

s₁: is the size of the nucleus (like the size of a grape)

d₁: is the distance between electrons and nucleus of example 1 = 1 mile

s₂: is the average diameter of the nucleus  = 10⁻¹³ cm

d₂: is the average distance between electrons and nucleus of example 2 = 10⁻⁸ cm

Assuming that the diameter of a grape is 3 cm (in a spherical way), the ratio of the <u>first example</u> is (eq 1):

r_{1} = \frac{3 cm}{1 mi*\frac{160934 cm}{1 mi}} = 1.86 \cdot 10^{-5}

Now, the ratio of the <u>second example</u> is (eq 2):

r_{2} = \frac{10^{-13} cm}{10^{-8} cm} = 1.00 \cdot 10^{-5}              

Since r₁ and r₂ are in the same order of magnitude (10⁻⁵), we can conclude that the given claim is reasonably accurate.      

You can learn more about the nucleus of an atom here: brainly.com/question/10658589?referrer=searchResults

I hope it helps you!                

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