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vitfil [10]
3 years ago
14

What does 3.5 degree of unsaturation tell you?

Chemistry
1 answer:
Kaylis [27]3 years ago
5 0

Answer:

It means the chemical entity is a radical

Explanation:

When we talk of unsaturation, we are referring to the number of pi-bonds in a chemical entity. The alkane, alkene and alkyne organic family are used to as common examples to explain the term unsaturation.

While alkynes have 3 bonds, it must be understood that they have 2 pi bonds only and as such their degree of saturation is two.

In the case of an alkene, there is only one single pi bond and as such the degree of unsaturation is 1.

Now in this case, we have a fractional 0.5 degree of unsaturation alongside the 3 to make a total of 3.5. So what’s the issue here?

The fractional part shows that the chemical entity we are dealing with here is a radical. While the integer 3 shows that there are 3 pi-bonds, the half pi bond remaining tells us that there is a missing electron on one of the atoms involved in the chemical bonding and as such, the 1/2 extra degree of unsaturation tends to tell us this.

Kindly recall that a radical is a chemical entity within which we have at the least an unpaired electron.

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How many moles of Al will be consumed when 0.400 mol of Al2O3 are produced in
Vladimir [108]

Answer:

C) 0.800 mol

Explanation:

  • 4Al + 3O₂ → 2Al₂O₃

In order to <u>convert from moles of Al₂O₃ into moles of Al</u>, we'll need to use<em> the stoichiometric coefficients</em>, using a conversion factor that has Al₂O₃ moles in the denominator and Al moles in the numerator:

  • 0.400 mol Al₂O₃ * \frac{4molAl}{2molAl_2O_3} = 0.800 mol Al

So the correct answer is option C).

7 0
3 years ago
How many moles are in 2.0 grams of Na?
lisov135 [29]

Answer:

46

Explanation:

Sodium metal has a molar mass of  

22.99

3 0
3 years ago
in an experiment 3.5g of element A reacted with 4.0g of element G to form a compound Calculate the empirical formula for this co
kolezko [41]

Additional information

Relative atomic mass(Ar) : A=7, G=16

The empirical formula : A₂G

<h3>Further explanation</h3>

Given

3.5g of element A

4.0g of element G

Required

the empirical formula for this compound

Solution

The empirical formula is the smallest comparison of atoms of compound forming elements.

The empirical formula also shows the simplest mole ratio of the constituent elements of the compound

mol of element A :

\tt mol=\dfrac{mass}{Ar}\\\\mol=\dfrac{3.5}{7}=0.5

mol of element G :

\tt mol=\dfrac{4}{16}=0.25

mol ratio A : G = 0.5 : 0.25 = 2 : 1

4 0
2 years ago
In an experiment, a 0.5297 g sample of diphenylacetylene (C14H10) is burned completely in a bomb calorimeter. The calorimeter is
V125BC [204]

Answer:

the Molar heat of  Combustion  of  diphenylacetylene (C_{14}H_{10})  = -6.931 *10^3 \ kJ/mol

Explanation:

Given that:

mass of diphenylacetylene (C_{14}H_{10}) = 0.5297 g

Molar Mass of diphenylacetylene (C_{14}H_{10}) = 178.21 g/mol

Then number of moles of diphenylacetylene (C_{14}H_{10})  = \frac{mass}{molar \ mass}

= \frac{0.5297  \ g }{178.24 \  g/mol}

= 0.002972 mol

By applying the law of calorimeter;

Heat liberated by 0.002972 mole of diphenylacetylene (C_{14}H_{10})  = Heat absorbed by H_2O + Heat absorbed  by the calorimeter

Heat liberated  by 0.002972 mole of diphenylacetylene (C_{14}H_{10})  =  msΔT + cΔT

= 1369 g  × 4.184 J g⁻¹°C⁻¹ × (26.05 - 22.95)°C + 916.9 J/°C (26.05 - 22.95)°C

= 17756.48 J + 2842.39 J

= 20598.87 J

Heat liberated by 0.002972 mole of diphenylacetylene (C_{14}H_{10})  = 20598.87 J

Heat liberated by 1 mole of  diphenylacetylene (C_{14}H_{10}) will be = \frac{20598.87 \ J}{0.002972 \ mol}

= 6930979.139 J/mol

= 6930.98 kJ/mol

Since heat is liberated ; Then, the Molar heat of  Combustion  of  diphenylacetylene (C_{14}H_{10})  = -6.931 *10^3 \ kJ/mol

3 0
3 years ago
Read 2 more answers
Round off the measurement to three significant figures 12.17º C
prohojiy [21]
12.2 C
It has 3 significant figures now.
6 0
3 years ago
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