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sleet_krkn [62]
2 years ago
5

Which of the following statements provides the BEST explanation for the many chemical similarities between two elements in a Mai

n Group? (Note: only ONE answer here.!) (a) They have similar atomic mass values. (b) Both have the same number of valence electrons. (c) They have the same number of protons. (d) They are in the same period.
Chemistry
1 answer:
rjkz [21]2 years ago
7 0

Answer:

(b) Both have the same number of valence electrons.  

Step-by-step explanation:

We find the most striking chemical similarities between two Main Group elements when they are in the same Group of the Periodic Table.

Elements in the same Group have the same number of valence electrons.

(a) is <em>wrong</em>, because elements in the same group have <em>different masses</em>.

(c) is <em>wrong,</em> because atoms with the same number of protons belong to the s<em>ame element</em>.

(d) is wrong, because elements in the same Group must be in . <em>different Periods.</em>

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The correct answer is option C, 5.02 x 10²² carbon atoms

Atomic mass of C = 12 g/mol

According to Avogadro, 1 mole of C has 6.023 x 10²³C atoms

Now 1 mole of C is equal to 12 g

Therefore, 12 g of C = 6.023 x 10²³ C atoms

1 g of C = \frac{6.023 x 10^2^3}{12} C atoms = 5.02 x 10²² C atoms

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2 years ago
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The number of grams of H2 in 1470 mL of H2 gas. ​
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Answer:

0.1313 g.

Explanation:

  • It is known that at STP, 1.0 mole of ideal gas occupies 22.4 L.
  • Suppose that hydrogen behaves ideally and at STP conditions.

<u><em>Using cross multiplication:</em></u>

1.0 mol of hydrogen occupies → 22.4 L.

??? mol of hydrogen occupies → 1.47 L.

∴ The no. of moles of hydrogen that occupies 1.47 L = (1.0 mol)(1.47 L)/(22.4 L) = 6.563 x 10⁻² mol.

  • Now, we can get the no. of grams of hydrogen in 6.563 x 10⁻² mol:

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A certain flexible weather balloon contains 7.4 L of helium gas. Initially, the balloon is in WP at 8500ft, where the temperatur
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<h2>The new volume of the balloon at the top of Pikes Peak is 10.2 L</h2>

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Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

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P_1 = initial pressure of gas = 577.0 torr

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V_1 = initial volume of gas = 7.4 L

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T_1 = initial temperature of gas = 20.6^oC=273+20.6=293.6K

T_2 = final temperature of gas = 7.5^oC=273+7.5=280.5K

Now put all the given values in the above equation, we get:

\frac{577.0\times 7.4}{293.6K}=\frac{400.0\times V_2}{280.5K}

V_2=10.2L

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