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Paladinen [302]

3 years ago

15

A team of researchers is investigating a new carbon-rich molecule important to life. They find evidence for the presence of both

cis and trans isomers of the molecule in a mixture of near-pure molecules. Which of the following must be contained in the new molecule?
A. A double bond
B. An aliphatic hydrocarbon
C. A hydrocarbon structure

1 answer:

ser-zykov [4K]3 years ago

6 0

Answer:

A. A double bond

Explanation:

Trans and cis Isomerism is a form of isomerism that specifically refers to alkenes, which is to say, aliphatic chains containing a <em>double bond</em> between two carbon atoms. Technically speaking, options B. and C. are both correct, because an alkene contains aliphatic hydrocarbons. However, neither of those options necessarily relate to cis and trans isomers.

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salantis [7]

The answer is B.) 10^5

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5 0

3 years ago

Read 2 more answers

For the following reactions, predict the products and write the balanced formula equation, complete ionic equation, and net ioni

stealth61 [152]

Answer:

.

Explanation:

To predict the products of these reactions we need to know the kind of reactions. All these reactions are double replacement reaction. In these kinds of reactions, the products will be the result of exchanging ions in the reactants. So, the first step is to identify the ions.

For the reaction, we have Hg2(NO3)2 and CuSO4. We have the ions Hg+1, NO3-1, Cu+2 and SO4-2

The way to make this exchange is putting together positive in one species with the negative of the other species. Following that rule we have

$Hg^{+1} - - - (SO_{4})^{-2}[/text]
the oxidation number will tell you the subscript for each species in the compound. In this case, is Hg2(SO4) [tex]Cu^{+2} - - - (NO_{3})^{-1} - - -> Cu(NO_{3})_{2} [/text] 
So, the products for this reaction will be
 [tex]Hg_{2} (NO_{3})_{2}(aq) + CuSO_{4}(aq) --> Hg_{2}SO_{4} + Cu(NO_{3})_{2}[/text]

After this, we proceed to balance the equation. For this, we check that we have the same number of each element on both sides of the equation. In this case, we can see that we have the same number, so the equation is balanced. Finally, we check the rules of solubility to see if the species are soluble in water or not. In this case sulfates area always soluble except for mercury so Hg2(SO4) precipitates in the solution (pre). Nitrates are always soluble so Cu(NO3)2 is soluble (aq) 
[tex] Hg_{2}(NO_{3})_{2}(aq) + CuSO_{4}(aq) - -> Hg_{2}SO_{4} (pre) + Cu(NO_{3})_{2}(aq)$

The complete ionic equation allows to show which of the reactants or products exist primarily as ions. For this reaction this will be:

$2Hg^{+1}(aq) + 2(NO_{3})^{-1}(aq) + (SO_{4})^{-2}(aq) + Cu^{+2}(aq) --> Hg_{2}SO_{4} (pre)+ Cu^{+2}(aq) + (NO_{3})^{-1}(aq) [/text]

To get net ionic equation we take away the ions that did not participate in the reactions. In other words the ones that are the same on both sides in the equation. In this case we see that [tex] Cu^{+2}(aq) and (NO_{3})^{-1}(aq) [/text] are the same on both sides so those ions are not include in the net ionic equation. This is:
[tex] 2Hg^{+1}(aq) + (SO_{4})^{-2}(aq) --> Hg_{2}SO_{4} (pre) [/text]

B [tex] Ni(NO_{3})_{2}(aq) + CaCl_{2}(aq)$

ions (1) $Ni^{+2} and (NO_{3})^{-1}$

ions (2) $Ca^{+2} and Cl^{-1}$

Exchanging

$Ni^{+2} ---- Cl^{-1} --> NiCl_{2}$

$Ca^{+2} --- (NO_{3})^{-1} --> Ca(NO_{3})_{2}$

Products

$Ni(NO_{3})_{2}(aq) + CaCl_{2}(aq) --> NiCl_{2} + Ca(NO_{3})_{2}$

The equation is already balanced

Chlorides are always soluble except Ag+, TI+, Pb+2 and Hg2+2. NiCl2 is soluble (aq)

Nitrates are always soluble. Ca(NO3)2 is soluble (aq)

Since both compounds are soluble, we can say that there is not reaction.

Complete ionic equation

$Ni^{+2}(aq) + 2(NO_{3})^{-1} (aq) + Ca^{+2}(aq) + 2Cl^{-1}(aq) - - > Ni^{+2}(aq) + 2(NO_{3})^{-1} (aq) + Ca^{+2}(aq) + 2Cl^{-1}(aq)$

Net ionic equation:

The ions in both sides of the equation are the same so all of them are cancelled and we cannot get a net ionic equation this explains why there is no reaction in this case.

C $K_{2}CO_{3}(aq) + MgI_{2}(aq)$

Ions(1) $K^{+1} and (CO_{3})^{-2}$

Ions(2) $Mg^{+2} and l^{-1}$

Exchanging

$K^{+1} --- l^{-1} - - > KI$

$Mg^{+2} --- (CO_{3})^{-2} - - > Ca(CO_{3})$

Products

$K_{2}CO_{3}(aq) + MgI_{2}(aq) - -> Kl + MgCO_{3}$

The equation is not balanced

Balance equation is

$K_{2}CO_{3}(aq) + MgI_{2}(aq) - -> 2Kl (aq) + MgCO_{3} (pre)$

iodides are always soluble except Ag+, TI+, Pb+2 and Hg2+2. KI is soluble (aq)

carbonates are always insoluble except group 1 cations. MgCO3 is insoluble (pre)

complete ionic equation

$2K^{+1}(aq) + (CO_{3})^{-2}(aq) + Mg^{+2}(aq) + 2l^{-1}(aq) - - > MgCO_{3} (pre) + 2K^{+1}(aq) + 2l^{-1}(aq)$

Net ionic equation

$(CO_{3})^{-2}(aq) + Mg^{+2}(aq) - - > MgCO_{3} (pre)$

D $Na_{2}CrO_{4}(aq) + AlBr_{3}(aq)$

Ions(1) $Na^{+1} and (CrO_{4})^{-2}$

Ions(2) $Al^{+3} and Br^{-1}$

Exchanging

$Na^{+1} ---- Br^{-1} - -> NaBr$

$Al^{+3} --- (CrO_{4})^{-2} - -> Al_{2}(CrO_{4})_{3}$

Products

$Na_{2}CrO_{4}(aq) + AlBr_{3}(aq) - -> NaBr + Al_{2}(CrO_{4})_{3}$

The equation is not balanced

Balance equation is

$3Na_{2}CrO_{4}(aq) + 2AlBr_{3}(aq) - -> 6NaBr + Al_{2}(CrO_{4})_{3}$

bromides are always soluble except Ag+, TI+, Pb+2 and Hg2+2. NaBr is soluble (aq)

chromates are always insoluble except group 1 cations. Al2(CrO4)3 is insoluble (pre)

$3Na_{2}CrO_{4}(aq) + 2AlBr_{3}(aq) - -> 6NaBr(aq) + Al_{2}(CrO_{4})_{3}(pre)$

Complete ionic equation

$6Na^{+1}(aq) + 3(CrO_{4})^{-2}(aq) + 2Al^{+3}(aq) + 6Br^{-1}(aq) - -> Al_{2}(CrO_{4})_{3}(pre) +6Br^{-1}(aq) + 6Na^{+1}(aq)$

Net ionic equation

$3(CrO_{4})^{-2}(aq) + 2Al^{+3}(aq) - -> Al_{2}(CrO_{4})_{3}(pre)$

6 0

3 years ago

Use the periodic table to identify the element with the electron configuration 1s²2s²2p⁴. Write its orbital diagram, and give th

natta225 [31]

Answer:

1. Orbital diagram

2p⁴ ║ ↑↓ ║ "↑" ║ ↑

2s² ║ ↑↓ ║

1s² ║ ↑↓ ║

2. Quantum numbers

<em>n </em>= 2,
<em>l</em> = 1,

$m_{l}$ = 0,

$m_{s}$ = +1/2

Explanation:

The fill in rule is:

Follow shell number: from the inner most shell to the outer most shell, our case from shell 1 to 2

Follow the The Aufbau principle, 1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<6s<4f<5d<6p<7s<5f<6d<7p
Hunds' rule: Every orbital in a sublevel is singly occupied before any orbital is doubly occupied. All of the electrons in singly occupied orbitals have the same spin (to maximize total spin).

So, the orbital diagram of given element is as below and the sixth electron is marked between " "

2p⁴ ║ ↑↓ ║ "↑" ║ ↑

2s² ║ ↑↓ ║

1s² ║ ↑↓ ║

The quantum number of an electron consists of four number:

<em>n </em>(shell number, - 1, 2, 3...)
<em>l</em> (subshell number or orbital number, 0 - orbital <em>s</em>, 1 - orbital <em>p</em>, 2 - orbital <em>d...</em>)
$m_{l}$ (orbital energy, or "which box the electron is in"). For example, orbital <em>p </em>(<em>l</em> = 1) has 3 "boxes", it was number from -1, 0, 1. Orbital <em>d</em> (<em>l </em>= 2) has 5 "boxes", numbered -2, -1, 0, 1, 2
$m_{s}$ (spin of electron), either -1/2 or +1/2

In our case, the electron marked with " " has quantum number

<em>n </em>= 2, shell number 2,
<em>l</em> = 1, subshell or orbital <em>p,</em>

$m_{l}$ = 0, 2nd "box" in the range -1, 0, 1

$m_{s}$ = +1/2, single electron always has +1/2

3 0

3 years ago

This condition affects more than 25 million Americans. Often triggered by allergies, this condition causes a constriction of the

gregori [183]

Common sense tells me that it is asthma cause i have it and yeah good luck

8 0

2 years ago

A molecule that oxidized gains electrons and energy. true or false

quester [9]

I suppose it false, since the oxidation involves the loss or removal of the electrons such forth it does not gain electrons.

5 0

3 years ago