What type of chemical bonding would exist between H-C-H?

Do you hace a copy of the student guide for this lab, including your date table?

kondor19780726 [428]

Answer:

yes i do

Explanation:

7 0

3 years ago

Read 2 more answers

Predict the carboxylic acid product of the following reaction:CH3COOCH2CH3ethyl ethanoate+H2OH+⟶carboxylic acid+CH3CH2OHethanolE

GaryK [48]

Answer: The IUPAC name of the carboxylic acid formed is ethanoic acid.

Explanation:

The basic rules for naming of organic compounds are :

First select the longest possible carbon chain.
The longest possible carbon chain should include the carbons of double or triple bonds.
The naming of alkane is done by adding the suffix -ane, alkene by adding the suffix -ene, alkyne by adding the suffix -yne and carboxylic acid by adding the suffix -oic acid.
The numbering is done in such a way that first carbon of double or triple bond gets the lowest number.
The carbon atoms of the double or triple bond get the preference over the other substituents present in the parent chain.

For the given chemical reaction, the equation follows:

$CH_3COOCH_2CH_3+H_2O\rightarrow CH_3COOH+CH_3CH_2OH$

When hydrolysis of ester takes place, it leads to the formation of an alcohol and a carboxylic acid.

So, when hydrolysis of ethyl ethanoate occurs, it produces ethanol and a 2- Carbon carboxylic acid named as ethanoic acid.

Hence, the IUPAC name of the carboxylic acid formed is ethanoic acid.

8 0

4 years ago

An argon ion laser emits light of wavelength of 488 nm. (a) Calculate the frequency of the light. (b) Suppose a pulse of light f

Alex787 [66]

Answer:

A) 6.15×10^14 per sec

B)2.5 sec

Explanation:

Wavelength of light = 488 nm

speed of light = 3×10^8 m/s

therefore, frequency of light $\nu=\frac{c}{\lambda}$

⇒ $\nu=\frac{3\times10^8}{488\times10^{-9}}$

⇒ν= 6.15×10^14 per sec

b) distance between moon and earth = 3.8×10^5 km.

= 3.8×10^8 m

light travel from earth moon and back to earth

therefore, distance traveled= 2× 3.8×10^8 m = 7.6×10^8 m

therefore time= distance/speed

⇒ time= $\frac{7.6\times10^8}{3\times10^8}$

= 2.5 sec

So, for a round trip time required is 2.5 sec.

5 0

3 years ago

Why is CCl4 a nonpolar molecule, but PCl3 is a polar molecule?

Anna11 [10]

If you follow the octet rule, you know that an element must have 8 outside (or valence) electrons to be energetically favorable.

In CCl4, the carbon molecule forms four bonds; one for each chlorine atom. Each bond contains 2 electrons, so it is satisfied.

In PCl3, Phosphorous forms only 3 bonds with chlorine, which means in order to have 8 valence electrons, it also has a lone pair of electrons, not bonded with chlorine.

Now, in CCl4, picture the shape of the molecule like a plus sign, with the carbon in the middle and the chlorine at the four ends. It is symmetrical, and therefore is nonpolar.

In PCl3, the lone pair electrons push, so to speak, the 3 chlorine atoms away, making a T-shaped molecule. Since the chlorine is more electronegative than carbon, the molecule is unbalenced, making it polar.

6 0

3 years ago

Using the bond energy data from your text (or the internet), determine (show calculations for) the approximate enthalpy change ,

aleksley [76]

Answer:

∆H= 438 KJ/mol

Explanation:

First, we have to find the energy bond values for each compound:

-) Cl-Cl = 243 KJ/mol

-) F-F = 159 KJ/mol

-) F-Cl = 193 KJ/mol

If we check the reaction we can calculate the number of bonds:

$Cl_2_(_g_)~+~3F_2_(_g_)~->~2ClF_3_(_g_)$

In total we will have:

-) Cl-Cl = 1

-) F-F = 3

-) F-Cl = 6

With this in mind. we can calculate the total energy for each bond:

-) Cl-Cl = (1*243 KJ/mol) = 243 KJ/mol

-) F-F = (3*159 KJ/mol) = 477 KJ/mol

-) F-Cl = (6*193 KJ/mol) = 1158 KJ/mol

Now, we can calculate the total energy of the products and the reagents:

Reagents = 243 KJ/mol + 477 KJ/mol = 720 KJ/mol

Products = 1158 KJ/mol

Finally, to calculate the total enthalpy change we have to do a subtraction between products and reagents:

∆H= 1158 KJ/mol-720 KJ/mol = 438 KJ/mol

I hope it helps!

7 0

3 years ago