Copper oxide is a: (Multiple Choice)

When a reaction occurs between atoms with ground state electron configuration 1s2 2s1 and 1s2 2s2 2p5 the predominant type of bo

fenix001 [56]

Answer: The bond formed between the elements will be ionic bond.

Explanation: We are given two elements having electronic configurations:

Element 1: $1s^22s^1$

Element 2: $1s^22s^22p^5$

Element 1 can easily loose 1 electron to attain stable electronic configuration and Element 2 can accept 1 electron to attain stable electronic configuration.

For these elements, there will be a complete transfer of electron from Element 1 to Element 2. Hence, this will form a ionic bond.

From the configuration, Element 1 is Lithium and Element 2 is Fluoride. So, the compound is LiF.

5 0

3 years ago

The compounds n-butane, ch3(ch2)2ch3, and trimethylamine, n(ch3)3, have very similar molecular weights. However, their melting p

Ostrovityanka [42]

<h3>Answer:</h3>

The lowest boiling point is of n-Butane because it only experiences London Dispersion Forces between molecules.

<h3>Explanation:</h3>

Lets take start with the melting point of both compounds.

n-Butane = - 140 °C

Trimethylamine = - 117 °C

Intermolecular Forces in n-Butane:

As we know n-Butane is made up of Carbon and Hydrogen atoms only bonded via single covalent bonds. The electronegativity difference between C and C atoms is zero while, that between C and H atoms is 0.35 which is less than 0.4. Hence, the bonds in n-Butane are purely non polar in nature. Therefore, only London Dispersion Forces are found in n-Butane which are considered as the weakest intermolecular interactions.

Intermolecular Forces in Trimethylamine:

Trimethylamine (a tertiary amine) is made up of Nitrogen, Carbon and Hydrogen atoms bonded via single covalent bonds. The electronegativity difference between N and C atoms is 0.49 which is greater than 0.4. Hence, the C-N bond is polar in nature. Therefore, Dipole-Dipole interactions will be formed along with London Dispersion Forces which are stronger than Dispersion Forces. Therefore, due to Dipole-Dipole interactions Trimethylamine will have greater melting point than n-Butane.

5 0

4 years ago

Does a reaction occur when aqueous solutions of silver(I) acetate and manganese(II) iodide are combined?

san4es73 [151]

Answer:

When aqueous solutions of silver(I) acetate and manganese(II) iodide are combined, an insoluble precipitate of silver(I) iodide is formed

Explanation:

It is an example of precipitation reaction.
When aqueous solutions of silver(I) acetate and manganese(II) iodide are combined, an insoluble precipitate of silver(I) iodide is formed.
Precipitation of silver(I) iodide is confirmed by it's yellow color.
Hence a reaction is observed.
Molecular reaction: $2Ag(OAc)(aq.)+MnI_{2}(aq.)\rightarrow 2AgI(s)+Mn(OAc)_{2}(aq.)$

7 0

3 years ago

While ethanol (CH3CH2OH) is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by

Anna [14]

<u>Answer:</u> The number of moles of ethanol after equilibrium is reached the second time is 11. moles.

<u>Explanation:</u>

We are given:

Initial moles of ethene = 34 moles

Initial moles of water vapor = 15 moles

The chemical equation for the formation of ethanol follows:

$CH_2=CH_2+H_2O\rightleftharpoons CH_3CH_2OH$

Initial: 34 15

At eqllm: 34-x 15-x x

We are given:

Equilibrium moles of ethene = 24 moles

Equilibrium moles of water vapor = 5 moles

Calculating for 'x'. we get:

$34-x=24\\\\x=10$

Volume of container = 100.0 L

The expression of $K_c$ for above equation follows:

$K_c=\frac{[CH_3CH_2OH]}{[CH_2=CH_2][H_2O]}$ .......(1)

$[CH_3CH_2OH]=\frac{10}{100}=0.1M$

$[CH_2=CH_2]=\frac{24}{100}=0.24M$

$[H_2O]=\frac{5}{100}=0.05M$

Putting values in expression 1, we get:

$K_c=\frac{0.1}{0.24\times 0.05}\\\\K_c=8.3$

Now, 11 moles of ethene gas is again added and equilibrium is re-established, we get:

The chemical equation for the formation of ethanol follows:

$CH_2=CH_2+H_2O\rightleftharpoons CH_3CH_2OH$

Initial: 24+11 5 10

At eqllm: 35-x 5-x 10+x

$[CH_3CH_2OH]=\frac{10+x}{100}$

$[CH_2=CH_2]=\frac{35-x}{100}$

$[H_2O]=\frac{5-x}{100}$

Putting values of in expression 1, we get:

$8.3=\frac{\frac{(10+x)}{100}}{\frac{(35-x)}{100}\times \frac{(5-x)}{100}}\\\\8.3=\frac{(10+x)\times 100}{(35-x)\times (5-x)}\\\\x^2-52x+55=0\\\\x=50.9,1.1$

The value of 'x' cannot exceed '35', so the numerical value of x = 50.9 is neglected.

Moles of ethanol = $(10+x)=10+1.1=11.$

Hence, the number of moles of ethanol after equilibrium is reached the second time is 11. moles.

8 0

3 years ago

when the relative humidity is 100%,we say the air is (3 pts) a) saturated b) supercooled c) superheated d) very humid

katovenus [111]

Answer:

The correct option is: a) saturated

Explanation:

Relative humidity is a primary measurement of humidity. At a given temperature, relative humidity describes the present state of absolute humidity relative to the maximum humidity.

It is generally expressed as percentage. Therefore, <u>100% relative humidity means that the air is entirely </u><u>saturated</u><u>.</u>

3 0

3 years ago