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2 years ago

What is the name of a compound with the structure: ch3ch2ch2ch2ch2co2ch2ch2ch3?

1 answer:

5 0

4-nonanone
Explanation: there is a functional group in the compound CH3CH2CH2CH2CH2COCH2CH2CH3 which is a ketone; So for naming a ketone we must first mention the name of the prefix that is correspond then add the suffix ''one''. And why I putted a ''4'' before the name is because we must indicate where this functional group is located, for doing that we have to start counting where the functional group is the closet to the end of the compound so when looking we should start to count on the right side. this is how I get 4-nonanone

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What is the composition, in atom percent, of an alloy that consists of a) 5.5 wt% Pb and b) 94.5 wt% of Sn? Assume that the atom

Anastaziya [24]

Answer : The percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

Explanation :

First we have to calculate the number of atoms in 5.5 wt% Pb and 94.5 wt% of Sn.

As, 207.2 g of lead contains $6.022\times 10^{23}$ atoms

So, 5.5 g of lead contains $\frac{5.5}{207.2}\times 6.022\times 10^{23}=1.59\times 10^{22}$ atoms

and,

As, 118.71 g of lead contains $6.022\times 10^{23}$ atoms

So, 94.5 g of lead contains $\frac{94.5}{118.71}\times 6.022\times 10^{23}=4.79\times 10^{23}$ atoms

Now we have to calculate the percent composition of Pb and Sn in atom.

$\% \text{Composition of Pb}=\frac{\text{Atoms of Pb}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Pb}=\frac{1.59\times 10^{22}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=3.21\%$

and,

$\% \text{Composition of Sn}=\frac{\text{Atoms of Sn}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Sn}=\frac{4.79\times 10^{23}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=96.8\%$

Thus, the percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

6 0

2 years ago

Using only the following elements P, Br, and Mg, give the formulas for:A. an ionic compound. B. a molecular compound with polar

Talja [164]

Answer:

<h2>1. Ionic compound- $MgBr_2$ </h2><h2>2. Polar molecular compound- $PBr_3$ </h2>

Explanation:

Mg is a metal that has 12 atomic numbers and thus its electronic configuration is $1s^22s^22p^63s^2$ . The outer most shell of this element has 2 electrons so it loses 2 electrons and thus form $Mg^2^+$ ions. Br is a nonmetal and has 35 atomic number so its electronic configuration is $1s^22s^22p^63s^23p^64s^23d^1^04p^5$ . Since its outermost shell has 7 electrons so it can accept one electron and thus forms $Br^-$ . So magnesium ion and bromide ion combine and forms an ionic compound $MgBr_2$ .

P is also a nonmetal and combine with Br with covalent bond and due to electronegativity differences form polar covalent compound such as $PBr_3$ .

8 0

2 years ago

Using the idea of electronegativity difference for predicting polarity of bonds and considering the shape of the following molec

Zinaida [17]

See electronegativity is the tendency of an atom to gain an electron and flourine with a valecy of one and a vey small size is the most electronegetive because its orbitals are quite closed to the nucleus and hence the attraction is quite strong so it can attract an electron.the question that arises is that some smaller atoms should be more electronegetive as they are closer to the nucleus but it need more energy for them as compared to flourine to complete their octet. now polarity increases when two atoms of quite different sizes form a compound ... the more electronegetive atom will always attract the bond pair forming a negetive charge on it and positive on the less electroneg. one and polarity increases with electronegetivity of the anion.now as your question says
<span>5=I2.. because both the atoms are same there wont be permanent polarity </span>
<span>4=HI iodine is the least electronegetive of all the halogens due to its large size,electronegetivity decreases down the group </span>
<span>3=HBr bromine is the 2nd largest halogen </span>
<span>2=HCl chlorine is the 3rd largest halogen </span>
<span>1=HF fluorine is the smallest halogen making and hence makes the most polar</span>

8 0

2 years ago

Iron-59 is used in medicine to diagnose blood circulation disorders. The half life of iron-59 is 44.5 days. How much of a 2.000

kolezko [41]

Answer:

0.258 mg of iron remains.

Explanation:

To solve this problem we can use the formula

M₂ = M₀ * $0.5^{\frac{t}{44.5} }$

Where M₂ is the mass remaining, M₀ is the initial mass, and t is time in days.

Using the data given by the problem:

M₂ = 2.000 mg * $0.5^{\frac{133.5}{44.5} }$

M₂ = 0.258 mg

7 0

2 years ago

Consider the balanced equation below. 4NH3 + 3O2 --> 2N2 + 6H2O What is the mole ratio of NH3 to N2?

AURORKA [14]

The balanced equation given is:
4NH3 + 3O2 .....> 2N2 + 6H2O

From this equation, we can note that 4 moles of NH3 are required to produce 2 moles of N2.

Therefore, the mole ratio of NH3 to N2 is 4:2 which can be simplified into 2:1

4 0

2 years ago

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