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3 years ago

How can I write ONE oxygen MOLECULE? pls help

Chemistry

1 answer:

omeli [17]3 years ago

4 0

One oxygen molecule is made of O2

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A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the

baherus [9]

Answer: 1. $9.08\times 10^{-6}$ moles

2. 90 mg

Explanation:

$O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus $4.54 \times 10^{-6}$ moles of ozone is removed by = $\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6}$ moles of sodium iodide.

Thus $9.08\times 10^{-6}$ moles of sodium iodide are needed to remove $4.54\times 10^{-6}$ moles of $O_3$

2. $\text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by = $\frac{2}{1}\times 0.0003=0.0006$ moles of sodium iodide.

Mass of sodium iodide= $moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg$ (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of $O_3$ .

3 0

3 years ago

Maintenance of............of the body such as temperature level, water content, pH and blood pressure to be in balanced and stab

n200080 [17]

This is homeostasis of the body I believe.

6 0

3 years ago

1. 2.0 mL/s x 4.0 hrs = ? L<br><br> How do you do this conversion problem

Phantasy [73]

Answer:

29 L.

Explanation:

Hello!

In this case, considering that we are performing a conversion by which the time should be cancelled out to obtain liters, we first need to convert the seconds on bottom to hours and then the volume on top to liters, just a shown down below:

$2.0\frac{mL}{s} *\frac{1L}{1000mL} *\frac{3600s}{1hr}*4.0hr\\\\=28.8L$

Which turns out 29 L with 2 significant figures.

Best regards!

6 0

2 years ago

Calculate the mass of 488 moles of calcium carbonate

san4es73 [151]

Answer: 48800g

Explanation:

Using the mathematical relation : Moles = Mass / Molar Mass

Moles = 488

Molar mass of CaCO3 = 40 + 12 + (16 x 3) = 100g/mol

Therefore

488 = mass / 100 = 48800g

6 0

3 years ago

student in chemistry 150-02 weighed out 55.5g of octane c8h18 and allowed it to react with oxygen, o2 the product formed were ca

Tasya [4]

Answer:

C8H8 + 10O2 → 8CO2 + 4H2O

Explanation:

unbalanced reaction:

C8H8 + O2 → CO2 + H2O

balanced for semireactions:

(1) 16H2O + C8H8 → 8CO2 + 40H+

(2) 10(4H+ + O2 → 2H2O)

⇒ 40H+ + 10O2 → 20H2O

(1) + (2):

balanced reaction:

⇒ C8H8 + 10O2 → 8CO2 + 4H2O

8 - C - 8

20 - O2 - 20

8 - H - 8

3 0

3 years ago