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3 years ago

Which of the following represents a hydrocarbon that contains one double bond somewhere in the carbon chain?

1 answer:

8 0

The answer is alkene.

An alkane is a saturated hydrocarbon, thi is it has only single bonds.

Alkenes and alkynes are unsaturated: alkenes have double bonds and alkynes have triple bonds.

Subsituted hydrocarbon, is a hydrocarbon with one hydrogen substituted by another element or a group.

For example:

CH3 - CH - CH3
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F

You might be interested in

Balance the following equation:

Firdavs [7]

Answer: a) $2K_2CrO_4+3Na_2SO_3+10HCl\rightarrow 4KCl+3Na_2SO_4+2CrCl_3+5H_2O$

b) 1 mole of $SO_2$ is produced.

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

The skeletal equation is:

$K_2CrO_4+Na_2SO_3+HCl\rightarrow KCl+Na_2SO_4+CrCl_3 +H_2O$

The balanced equation will be:

$2K_2CrO_4+3Na_2SO_3+10HCl\rightarrow 4KCl+3Na_2SO_4+2CrCl_3+5H_2O$

Thus the coefficients are 2, 3 , 10 , 4 , 3 , 2 and 5.

b) Oxidation: $2I-^-\rightarrow I_2+2e-^-$

Reduction: $SO_4^{2-}+2e^-+4H^+\rightarrow SO_2+2H_2O$

Net reaction: $2I-^-+SO_4^{2-}+4H^+\rightarrow I_2+SO_2+2H_2O$

When 1 mole of $I_2$ is produced, 1 mole of $SO_2$ is produced.

8 0

3 years ago

Plzz help me out guys ......10 points rewarded.....!!!!!

GaryK [48]

number 5 is 1 : 1

number1 is also 1:1 ibelieve but it could be 2:1 just like number 5 but im postive its 1:1 for nummer 1 and number 5

4 0

3 years ago

What is the maximum number of moles of N-acetyl-p-toluidine can be prepared from 70. milliliters of 0.167 M p-toluidine hydrochl

Kitty [74]

Answer:

$\large \boxed{\text{0.012 mol}}$

Explanation:

We will need a balanced equation with moles, so let's gather all the information in one place.

CH₃C₆H₄NH₂·HCl + (CH₃CO)₂O ⟶ CH₃C₆H₄NHCOCH₃ + junk

V/mL: 70.

c/mol·L⁻¹: 0.167

For simplicity in writing , let's call p-toluidine hydrochloride A and N-acetyl-<em>p</em>-toluidine B.

The equation is then

A + Ac₂O ⟶ B + junk

1. Moles of A

$\text{Moles of A} = \text{70. mL A}\times \dfrac{\text{0.167 mmol A}}{\text{1 mL A}}= \text{12 mmol A}$

2. Moles of B

The molar ratio is 1 mol B:1 mol A

Moles of B = moles of A = 12 mmol = 0.012 mol

$\text{You can prepare $\large \boxed{\textbf{0.012 mol}}$ of N-acetyl-p-toluidine. }$

3 0

2 years ago

Stannum has a body centered tetragonal with lattice constant, a = b = 5.83A and c = 3.18A. If the atomic radius is 0.145 nm, det

Fed [463]

Answer:

the atomic packing factor of Sn is 0.24

Explanation:

a = b = 5.83A and c = 3.18A.

Volume of unit cell = a²c

= (5.83)² * 3.18 * 10⁻²⁴ cm³

= 1.08 * 10⁻²²cm³

Volume of atoms =

$2 \times \frac{4}{3} \pi r^3$

(∴ BCC, effective number of atom is 2)

Volume of atoms =

$2 * \frac{4}{3} *3.14*(0.145*10^-^7cm)^3$

= 2.55*10⁻²³cm³

$\text {Atomic packing factor}=\frac{\text {volume occupied by atom}}{\text {volume of unit cell }}$

$=\frac{2.55*10^-^2^3}{1.08*10^-^2^2} \\\\=0.24$

<h3>therefore, the atomic packing factor of Sn is 0.24</h3>

4 0

3 years ago

Using the standard enthalpies of formation found in the textbook, determine the enthalpy change for the combustion of ethanol c2

ArbitrLikvidat [17]

Enthalpy of formation is calculated by subtracting the total enthalpy of formation of the reactants from those of the products. This is called the HESS' LAW.
ΔHrxn = ΔH(products) - ΔH(reactants)

Since the enthalpies are not listed in this item, from reliable sources, the obtained enthalpies of formation are written below.
ΔH(C2H5OH) = -276 kJ/mol
ΔH(O2) = 0 (because O2 is a pure substance)
ΔH(CO2) = -393.5 kJ/mol
ΔH(H2O) = -285.5 kJ/mol

Using the equation above,
ΔHrxn = (2)(-393.5 kJ/mol) + (3)(-285.5 kJ/mol) - (-276 kJ/mol)
ΔHrxn = -1367.5 kJ/mol

<em>Answer: -1367.5 kJ/mol</em>

6 0

2 years ago

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