Which molecule will have a double covalent bond? #3 #5 & #6

What does 3.5 degree of unsaturation tell you?

Kaylis [27]

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.

5 0

3 years ago

How many amino acids can compose a protein molecule?

Vladimir79 [104]

The answer is A: Between 50 and 5,000 amino acids

3 0

2 years ago

how many kilograms of a 35% m/m sodium chlorate solution is needed to react completely with 0.29 l of a 22% m/v aluminum nitrate

Stolb23 [73]

Answer:- 0.273 kg

Solution:- A double replacement reaction takes place. The balanced equation is:

$3NaClO_3+Al(NO_3)_3\rightarrow 3NaNO_3+Al(ClO_3)_3$

We have 0.29 L of 22% m/v aluminum nitrate solution. m/s stands for mass by volume. 22% m/v aluminium nitrate solution means 22 g of it are present in 100 mL solution. With this information, we can calculate the grams of aluminum nitrate present in 0.29 L.

$0.29L(\frac{1000mL}{1L})(\frac{22g}{100mL})$

= 63.8 g aluminum nitrate

From balanced equation, there is 1:3 mol ratio between aluminum nitrate and sodium chlorate. We will convert grams of aluminum nitrate to moles and then on multiplying it by mol ratio we get the moles of sodium chlorate that could further be converted to grams.

We need molar masses for the calculations, Molar mass of sodium chlorate is 106.44 gram per mole and molar mass of aluminum nitrate is 212.99 gram per mole.

$63.8gAl(NO_3)_3(\frac{1mol}{212.99g})(\frac{3molNaClO_3}{1molAl(NO_3)_3})(\frac{106.44g}{1mol})$

= $95.7gNaClO_3$

sodium chlorate solution is 35% m/m. This means 35 g of sodium chlorate are present in 100 g solution. From here, we can calculate the mass of the solution that will contain 95.7 g of sodium chlorate and then the grams are converted to kg.

$95.7gNaClO_3(\frac{100gSolution}{35gNaClO_3})(\frac{1kg}{1000g})$

= 0.273 kg

So, 0.273 kg of 35% m/m sodium chlorate solution are required.

7 0

3 years ago

? Answer the question below. Type your response in the space provided. What volume of a 2.5 M stock solution of acetic acid (HC2

Novay_Z [31]

Data:
$M_{concentrated} = 2.5\:mol$
$V_{concentrated} = ?$
$M_{dilute} = 0.50\:mol$
$V_{dilute} = 100\:mL\to0.100\:L$

Formula: Dilution Calculations

 $M_{concentrated} * V_{concentrated} = M_{dilute} * V_{dilute}$

Solving:


$M_{concentrated} * V_{concentrated} = M_{dilute} * V_{dilute}$
$2.5 * V_{concentrated} = 0.50 * 0.100$
$2.5V_{concentrated} = 0.05$
$V_{concentrated} = \frac{0.05}{2.5}$
$\boxed{\boxed{V_{concentrated} = 0.02\:L\:or\:20\:mL}} \end{array}}\qquad\quad\checkmark$

3 0

3 years ago

When an object is lifted 6 meters off the ground, it gains a certain amount of potential energy. If the same object is lifted 18

AlexFokin [52]

Answer:

The answer to your question is the letter C. three times as much

Explanation:

Data

First step = 6 m

Second step = 18 m

Potential energy is the energy stored that depends on its position.

Formula

Pe = mgh

m = mass; g = gravity; h = height

Potential energy of the first step

Pe1 = 6mg

Potential energy of the second step

Pe2 = 18mg

-Divide the Pe2 by the Pe1

Pe2/Pe1 = 18mg/6mg

= 3

7 0

2 years ago