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3 years ago

BF3 type of compound

Chemistry

1 answer:

USPshnik [31]3 years ago

6 0

Answer:

Corresponds to a covalent compound.

Explanation:

Both Boron and Fluorine are nonmetals, forming a covalent bond, the result of which is the boron trifluoride compound (BF3). This is a highly toxic gas.

It is synthesized by the reaction between hydrogen fluoride and boron oxide:

B2O3 + 6 HF → 2 BF3 + 3 H2O

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If 5 grams of co and 5 grams of o2 are combined according to the reaction 2co o2 --> 2co2, which is the limiting reagent?

IrinaK [193]

When 5 grams of CO and 5grams of O₂ is combined according to the reaction 2CO + O₂ ----------> 2CO₂ then Limiting reagent is CO.

Limiting reagent is the reactant that get used up in the reaction first.

According to the given reaction:

2CO + O₂ ----------> 2CO₂

5g 5g

∴ Molar mass of CO = Molar mass of C + Molar mass of O

⇒ Molar mass of CO = 14 + 16

⇒ Molar mass of CO = 28g

∴ Molar mass of O₂ = 16(2) = 32g

∴ Molar mass of CO₂ = Molar mass of C + 2(Molar mass of O)

⇒ Molar mass of CO₂ = 14 + 2(16)

⇒Molar mass of CO₂ =44g

Let's find out the moles of CO and O₂

∴ Moles = Given mass / Molar mass

⇒ moles of CO = 5/28 = 0.17

⇒ moles of O₂ = 5/32 = 0.15

For finding out the Limiting Reagent, we will divide the number of moles with the stiochiometry of the given reaction.

⇒ For CO = Moles/ stiochiometry = 0.17/2 = 0.085

⇒ For O₂ = Moles/ stiochiometry = 0.15/1 = 0.15

Since, the ratio of number of moles with the stiochiometry is less for CO hence it is the Limiting reagent, i.e. it will get used up in the reaction first.

Hence, the Limiting reagent for the reaction is CO.

Learn more about Limiting reagent here, brainly.com/question/11848702

#SPJ4

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malfutka [58]

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Andrej [43]

The Molarity of the solution is given by:
$Molarity = \frac{moles-of-solute}{liters-of-solution}$ --- (A)

But NaCl, which is a solute, is given in grams.
Therefore, first, we need to convert grams into moles.

Molecular weight of NaCl = 58.44 grams/mole.

So,
117 grams * (moles/ 58.44 grams) = 2.00 moles of NaCl

Now that we have moles of NaCl, plug in the values in (A):
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The correct answer is: 0.741 moles/litre = Molarity

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Answer:

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