The best and most correct answer among the choices provided by your question is the fourth choice.
Copper <span>has an average atomic mass of about 63.5 </span>amu<span>.</span>
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D. Volcanoes formed most of them through eruption.
Answer:
A
Explanation:
Same air has special carakteristik features
<u>Answer:</u>
<u>For 2:</u> The % yield of the product is 92.34 %
<u>For 3:</u> 12.208 L of carbon dioxide will be formed.
<u>Explanation:</u>
The percent yield of a reaction is calculated by using an equation:
......(1)
Given values:
Actual value of the product = 78.4 g
Theoretical value of the product = 84.9 g
Plugging values in equation 1:

Hence, the % yield of the product is 92.34 %
The number of moles is defined as the ratio of the mass of a substance to its molar mass.
The equation used is:
......(2)
Given mass of carbon dioxide = 24 g
Molar mass of carbon dioxide = 44 g/mol
Plugging values in equation 1:

<u>At STP conditions:</u>
1 mole of a gas occupies 22.4 L of volume
So, 0.545 moles of carbon dioxide will occupy =
of volume
Hence, 12.208 L of carbon dioxide will be formed.
Answer:
3.18 (w/w) %
Explanation:
In the problem, you can find mass of NaClO knowing the reaction of NaClO with Na₂S₂O₃ is:
NaClO + 2Na₂S₂O₃ + H₂O → NaCl + Na₂S₄O₆ +2NaOH + NaCl
<em>Where 1 mole of NaClO reacts with 2 moles of Na₂S₂O₃</em>
<em> </em>Moles of thiosulfate in the titration are:
0.0101L ₓ (0.042mol / L) = 4.242x10⁻⁴ moles of Na₂S₂O₃
Thus, moles of NaClO in the initial solution are:
4.242x10⁻⁴ moles of Na₂S₂O₃ ₓ (1mol NaClO / 2 mol Na₂S₂O₃) = 2.121x10⁻⁴ moles NaClO
As molar mass of NaClO is 74.44g/mol, mass of 2.121x10⁻⁴ moles are:
2.121x10⁻⁴ moles ₓ (74.44g / mol) = <em>0.0158g of NaClO</em>
As mass of bleach is 0.496g, mass percent is:
0.0158g NaClO / 0.496g bleach ₓ 100 =
<h3>3.18 (w/w) % </h3>