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3 years ago

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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..

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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Suppose that 0.410 mol of methane, CH4(g), is reacted with 0.560 mol of fluorine, F2(g), forming CF4(g) and HF(g) as sole produc

Ber [7]

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

1 mole of CH4 → 4 mole of 4 mole of fluorine

0.41 mole of methane → 4*0.41 = 1.64 mole of fluorine for complete reaction

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.

4 moles of fluorine → 1 mole of CF4

0.56 mole → $\frac{1}{4} * 0.56$ = 0.14mole of CF4

4 moles of fluorine → 4 moles of HF

0.56 mole of fluorine → 0.56 mole of HF

now to find the heat released we have the formula as

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

where n is the moles of the reactant and product.

note: since no information is given about the enthalpies of the species we leave it on general equation also you need to add the product side enthalpy of the species present and similarly on the product side.

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3 years ago

Dioxin, produced as a by-product of various industrial chemical processes, is suspected of contributing to the development of ca

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Answer:

E

Explanation:

Why can a signaling molecule cause different responses in different cells? Different cells have membrane receptors that bind to different sides of the signaling molecule. The transduction process is unique to each cell type; to respond to a signal, different cells require only a similar membrane receptor

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Light travels as a/an___________

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Answer:

D. Electromagnetic

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Read 2 more answers

Answer the following for the reaction: 3AgNO3(aq)+Na3PO4(aq)→Ag3PO4(s)+3NaNO3(aq)

Brums [2.3K]

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

0.62577 grams of $Ag_3PO_4$ is produced.

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Explanation:

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Lead can pose a significant risk to your health if too much of it enters your body.
Lead enters drinking water primarily as a result of the corrosion.
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Lead in drinking rarely cause of lead poisoning can significantly increase a person’s total lead exposure.
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EPA estimates that drinking water can make up to 20% or more of a person’s total exposure to lead which is dangerous.

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