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galben [10]
3 years ago
13

Plz answer this question.....I know the answer for this question but I want to know the explanation plz plz plz I want to know t

he reason for the answer...plz explain properly.....and I also wanted to ask how can we know if a compound is double,triple,single bonded?????
And also how to find the alkene and alkane of a compund if the alkyne is given ...plz answer I need your help .... urgent....I will mark the answer as brainliest..

Chemistry
1 answer:
vlabodo [156]3 years ago
8 0

The compound C_{3}H_{4} will have a triple bond.

Explanation:

A compound which consists of carbon and hydrogen atoms is known as a hydrocarbon.

Alkanes, alkenes and alkynes are all hydrocarbons.

  • General chemical formula of an alkane is C_{n}H_{2n+2}. In an alkane molecule, all the atoms will be bonded through single bonds.

For example, C_{3}H_{8} is propane.

  • General chemical formula of an alkene is C_{n}H_{2n}. An alkene molecule will have atleast one double bond between two carbon atoms.

For example, C_{2}H_{4} is ethene.

  • General chemical formula of an alkyne is C_{n}H_{2n-2}. An alkyne will have atleast one triple bond between two carbon atoms.

For example, C_{3}H_{4} is propyne.

Thus, we can conclude that out of the given options C_{3}H_{4} will have a triple bond.

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Answer:

The balanced equation for this reaction will be

                            CH4 + 4F2    →  CF4 + 4HF

We can see that 1 mole of methane requires 4 moles of fluorine but we have 0.41 moles of CH4 and 0.56mole of F2

So using the unitary method we will get that

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but we have only 0.56 mole of fluorine that means fluorine is the limiting reagent and the product will only be formed by only this amount of fluorine.

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now to find the heat released we have the formula as

DELTA H = n * Delta H of product - n *delta H of reactant

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Answer the following for the reaction: 3AgNO3(aq)+Na3PO4(aq)→Ag3PO4(s)+3NaNO3(aq)
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Answer:1) Volume of AgNO_3 required is 55.98 mL.

2) 0.62577 grams of Ag_3PO_4 is produced.

Explanation:

3AgNO_3(aq)+Na_3PO_4(aq)\rightarrow Ag_3PO_4(s)+3NaNO_3(aq)

1) Molarity of AgNO_3,M_1=0.225 M

Volume of AgNO_3.V_1=?

Molarity of Na_3PO_4,M_2=0.135 M

Volume of Na_3PO_4,V_2=31.1 mL=0.0311 L

Molarity=\frac{\text{number of moles}}{\text{volume of solution in liters}}

\text{number of moles }Na_3PO_4=M_2\times V_2=0.135 mol/L\times 0.0311 L=0.0041985 moles

According to reaction, 1 mole of Na_3PO_4 reacts with 3 mole of AgNO_3, then, 0.0041985 moles of Na_3PO_4 will react with:

\frac{3}{1}\times 0.0041985 moles of AgNO_3 that is 0.0125955 moles.

M_1=0.225 M=\frac{\text{number of moles of }AgNO_3}{V_1}

V_1=\frac{0.0125955 moles}{0.225 M}=0.05598 L=55.98 mL

Volume of AgNO_3 required is 55.98 mL.

2)

Molarity=0.195 M=\frac{\text{number of moles}}{\text{volume of solution in liters}}

Number of moles of AgNO_3=0.195\times 0.023 L=0.004485 moles

According to reaction, 3 moles of AgNO_3 gives 1 mole of Ag_3PO_4, then 0.004485 moles of AgNO_3 will give:\frac{1}{3}\times 0.004485 moles of Ag_3PO_4 that is 0.001495 moles.

Mass of Ag_3PO_4 =

Moles of Ag_3PO_4 × Molar Mass of Ag_3PO_4

= 0.001495 moles × 418.58 g/mol = 0.62577 g

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