B. reproduction doesn’t require mate
The amount, in mg, of CO present in the room will be 191,520 mg.
<h3>Stoichiometric problem</h3>
The concentration of the gas in the room is 5.7 x 
mg/cm3.
The dimension of the room is 3.5 m x 3.0 m x 3.2 m. This is equivalent to 350 cm x 300 cm x 320 cm.
We can obtain the volume of the room as:
350 x 300 x 320 = 33,600,000 cm3
The concentration is in mg/cm3, meaning that it is mass/volume.
Thus:
mass = concentration x volume = 5.7 x mg/cm3 x 33,600,000 cm3
= 191,520 mg
The mass of CO in the room is 191,520 mg
More on stoichiometric problems can be found here: brainly.com/question/14465605
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B. The energy barrier between reactants and products
hope this helps!
(:
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>
The best option is a. but it is not complete!
Both liquids have different densities, it's true. But they are immiscible liquids and they don't form a solution.
