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3 years ago

The formula B+ O2_ B2O3 is balanced as

Chemistry

1 answer:

julsineya [31]3 years ago

5 0

Hey there!

B + O₂ → B₂O₃

Balance O.

2 on the left, 3 on the right. Add a coefficient of 3 in front of O₂ and a coefficient of 2 in front of B₂O₃.

B + 3O₂ → 2B₂O₃

Balance B.

1 on the left, 4 on the right. Add a coefficient of 4 in front of B.

4B + 3O₂ → 2B₂O₃

Our final balanced equation: 4B + 3O₂ → 2B₂O₃

Hope this helps!

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grandymaker [24]

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3 years ago

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38.2mm converted into cm

Alex

1 mm ---------- 0.1 cm
38.2 mm ------ ?

38.2 x 0.1 / 1

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4 years ago

Calculate the number of milliliters of 0.440 M KOH required to precipitate all of the Fe2+ ions in 187 mL of 0.692 M FeSO4 solut

EleoNora [17]

Answer:

588.2 mL

Explanation:

FeSO₄(aq) + 2KOH(aq) → Fe(OH)₂(s) + K₂SO₄(aq)

First we calculate how many Fe⁺² moles reacted, using the given concentration and volume of FeSO₄ solution (the number of FeSO₄ moles is equal to the number of Fe⁺² moles):

moles = molarity * volume

187 mL * 0.692 M = 129.404 mmol Fe⁺²

Then we convert Fe⁺² moles to KOH moles, using the stoichiometric ratios:

129.404 mmol Fe⁺² * $\frac{2mmolKOH}{1mmolFeSO_4}$ = 258.808 mmol KOH

Finally we calculate the required volume of KOH solution, using the given concentration and the calculated moles:

volume = moles / molarity

258.808 mmol KOH / 0.440 M = 588.2 mL

6 0

3 years ago

For each solute identify the better solvent: water or carbon tetrachloride Na2, I2, CCH2O, C6H6

Virty [35]

NH3 = water (it is actually a water soluble gas)
C6H14 = CCl4 (C6H14 won't mix with water at all)
Na2S = water (Na2S is a salt. Salts dissolve best in water)
Br2 = CCl4, but it will also dissolve in water.

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3 years ago

Assuming the same temperature and pressure for each gas, how many milliliters of carbon dioxide are produced from 16 0 mL of CO

aivan3 [116]

Answer:

$V_{CO_2}=16.0mL$

Explanation:

Hello,

In this case, given that the same temperature and pressure is given for all the gases, we can notice that 16.0 mL are related with two moles of carbon monoxide by means of the Avogadro's law which allows us to understand the volume-moles relationship as a directly proportional relationship. In such a way, since in the chemical reaction:

$2CO(g)+O_2(g)\rightarrow 2CO_2(g)$

We notice two moles of carbon monoxide yield two moles of carbon dioxide, therefore we have the relationship:

$n_{CO}V_{CO}=n_{CO_2}V_{CO_2}$

Thus, solving for the yielded volume of carbon dioxide we obtain:

$V_{CO_2}=\frac{n_{CO}V_{CO}}{n_{CO_2}} =\frac{2mol*16.0mL}{2mol}\\ \\V_{CO_2}=16.0mL$

Best regards.

7 0

3 years ago