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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

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How much greater is the pressure at the deepest part of the ocean compared to that found at sea level?

Leno4ka [110]

The deepest part of the ocean is the Marianas Trench, where the water pressure is 1,000 times that of which is found at sea level.

6 0

3 years ago

Read 2 more answers

Rank the following atoms in order of increasing electronegativity.a. Se, O, S b. P, Na, Cl c. Cl, S, F d. O, P, N

Papessa [141]

Answer:

a)- Se, S, O

b)- Na, P, Cl

c)- S, Cl, F

d)- P, N, O

Explanation:

We can solve this problem by looking at the Periodic Table. In a group, electronegativity increases from bottom to top; whereas in periods it increases from left to right.

Thus, we order the elements from lower to higher electronegativity as follows:

a. Se, O, S ⇒ order: Se, S, O

Because they are all in the same group. Se is near the bottom, followed by S and O is at the top.

b. P, Na, Cl ⇒ order: Na, P, Cl

They are in the same period. Na is at the left, followed by P and Cl is nearest the right.

c. Cl, S, F ⇒ order: S, Cl, F

P and Cl are in the same period, and P is at the left, so it has the lowest electronegativity. F is in the same group of Cl, but at the top. F has the highest electronegativity.

d. O, P, N ⇒ order: P, N, O

N and P are in the same group, but P is at the bottom so it has the lower electronegativity. N and O are in the same period, but O is at the right, so it is the most electronegative.

8 0

3 years ago

The chemical bonds of carbohydrates and lipids have high potential energy because:

gladu [14]

Answer:

c. Many of their bonds are C-C and C-H

Explanation:

The majority of bonds in carbohydrates and lipids( being an organic compound) are C-C and C-H. Like glucose, fructose or galactose ,etc.

These bonds are strong and do require a lot of energy to break. Thus, a lot of energy are required to break carbs and lipids into simpler compounds.Therefore, carbohydrates and lipids have high potential energy.

The correct answer is c.

4 0

3 years ago

How much heat is released when 75 g of octane is burned completely if the enthalpy of combustion is -5,500 kJ/mol CsH18? C8H18 +

aalyn [17]

Answer : The correct option is, (D) 3600 kJ

Explanation :

Mass of octane = 75 g

Molar mass of octane = 114.23 g/mole

Enthalpy of combustion = -5500 kJ/mol

First we have to calculate the moles of octane.

$\text{ Moles of octane}=\frac{\text{ Mass of octane}}{\text{ Molar mass of octane}}=\frac{75g}{114.23g/mole}=0.656moles$

Now we have to calculate the heat released in the reaction.

As, 1 mole of octane released heat = -5500 kJ

So, 0.656 mole of octane released heat = 0.656 × (-5500 kJ)

= -3608 kJ

≈ -3600 kJ

Therefore, the heat released in the reaction is 3600 kJ

5 0

3 years ago

Five million gallons per day (MGD) of wastewater, with a concentration of 10.0 mg/L of a conservative pollutant, is released int

hjlf

Answer:

a) The concentration in ppm (mg/L) is 5.3 downstream the release point.

b) Per day pass 137.6 pounds of pollutant.

Explanation:

The first step is to convert Million Gallons per Day (MGD) to Liters per day (L/d). In that sense, it is possible to calculate with data given previously in the problem.

Million Gallons per day $1 MGD = 3785411.8 litre/day = 3785411.8 L/d$

$F_1 = 5 MGD (\frac{3785411.8 L/d}{1MGD} ) = 18927059 L/d\\F_2 =10 MGD (\frac{3785411.8 L/d}{1MGD} )= 37854118 L/d$

We have one flow of wastewater released into a stream.

First flow is F1 =5 MGD with a concentration of C1 =10.0 mg/L.

Second flow is F2 =10 MGD with a concentration of C2 =3.0 mg/L.

After both of them are mixed, the final concentration will be between 3.0 and 10.0 mg/L. To calculate the final concentration, we can calculate the mass of pollutant in total, adding first and Second flow pollutant, and dividing in total flow. Total flow is the sum of first and second flow. It is shown in the following expression:

$C_f = \frac{F1*C1 +F2*C2}{F1 +F2}$

Replacing every value in L/d and mg/L

$C_f = \frac{18927059 L/d*10.0 mg/L +37854118 L/d*10.0 mg/L}{18927059 L/d +37854118 L/d}\\C_f = \frac{302832944 mg/d}{56781177 L/d} \\C_f = 5.3 mg/L$

a) So, the concentration just downstream of the release point will be 5.3 mg/L it means 5.3 ppm.

Finally, we have to calculate the pounds of substance per day (Mp).

We have the total flow F3 = F1 + F2 and the final concentration $C_f$ . It is required to calculate per day, let's take a time of t = 1 day.

$F3 = F2 +F1 = 56781177 L/d \\M_p = F3 * t * C_f\\M_p = 56781177 \frac{L}{d} * 1 d * 5.3 \frac{mg}{L}\\M_p = 302832944 mg$

After that, mg are converted to pounds.

$M_p = 302832944 mg (\frac{1g}{1000 mg} ) (\frac{1Kg}{1000 g} ) (\frac{2.2 lb}{1 Kg} )\\M_p = 137.6 lb$

b) A total of 137.6 pounds pass a given spot downstream per day.

4 0

3 years ago