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3 years ago

Which of the following compounds contains a double bond? A - NH3 B - CO2 C - N2 D - CI2

Chemistry

1 answer:

notka56 [123]3 years ago

6 0

Answer:- B) $CO_2$

Explanations:- In ammonia, N is the central atom and three H atoms are bonded to it via single bonds.

In carbon dioxide, the central atom is C and two O atoms are bonded to it via double bonds. The reason is, carbon has four valence electrons and oxygen has 6 valence electrons. To complete the octet, each oxygen makes a double bond with carbon. It looks as O=C=O .

Nitrogen has 5 valence electrons and to complete its octet, it needs 3 more electrons so both N atoms are bonded to each other via a triple bond.

Cl has 7 valence electrons and needs one more to complete its octet. So, there is a single bond between two Cl atoms.

Hence the right choice is B) $CO_2$ .

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18 the first step in the mechanism is the acid catalyzed generation of an enol and then electrophilic addition of bromine. which

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the first step in the mechanism is the acid-catalyzed generation of an enol and then electrophilic addition of bromine and cation is formed because of the destabilization effect of the electronegativity of oxygen

The ability of an atom or functional group to draw electrons to itself is known as an electronegativity in chemistry. An atom's electronegativity is influenced by both its atomic number and how far away from its charged nuclei its valence electrons are located.

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The propensity of an atom to attract other atoms when it is combined is known as an element's electronegativity. Additionally, a pair of bound electrons are shared. In contrast, an element's electropositivity refers to an atom's propensity to contribute electrons while also withdrawing from covalent connections.

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2 years ago

This drawing shows the internal anatomy of a sponge.

Evgen [1.6K]

Answer:

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Explanation:

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3 years ago

How many of the following are found in 15.0 kmol of xylene (C8H10)? (a) kg C8H10; (b) mol C8H10; (c) lb-mole C8H10; (d) mol (g-a

77julia77 [94]

Answer:

a) 1592.4 kg C8H10

b) 15*10³ mol C8H10

c) 33.1 lb-mole

d) 1.2 *10^5 mol C

e) 1.5 * 10^5 mol H

f) 1.44 * 10^6 grams C

g) 1.52 * 10^5 grams H

h) 9.03*10^27 molecules C8H10

Explanation:

Step 1: Data given

Number of moles C8H10 = 15.0 kmol =15000 moles

Molar mass of C8H10 = 106.16 g/mol

Step 2: Calculate mass C8H10

Mass C8H10 = moles C8H10 * molar mass C8H10

Mass C8H10 = 15000 * 106.16 g/mol

Mass C8H10 = 1592400 grams = 1592.4 kg

Step 3: Calculate moles C8H10

15.0 kmol = 15*10³ mol C8H10

Step 4: Calculate lb-mol

15.0 kmol = 33.1 lb-mole

Step 5: Calculate moles of C

For 1 mol C8H10 we have 8 moles of C

For 15*10³ mol C8H10 we have 8* 15*10³ =1.2 *10^5 mol C

Step 6: Calculate moles H

For 1 mol C8H10 we have 10 moles of H

For 15*10³ mol C8H10 we have 10* 15*10³ = 1.5 * 10^5 mol H

Step 7: Calculate mass of C

Mass C = moles C * molar mass C

Mass C = 1.2 *10^5 mol C * 12 g/mol

Mass C = 1.44 * 10^6 grams

Step 8: Calculate mass of H

Mass H = moles H * molar mass H

Mass H = 1.5 *10^5 mol H * 1.01 g/mol

Mass C = 1.52 * 10^5 grams

Step 9: Calculate molecules of C8H10

Number of molecules = number of moles * number of Avogadro

Number of molecules = 15*10³ mol C8H10 * 6.022*10^23

Number of molecules = 9.03*10^27 molecules

8 0

3 years ago

A cube is 4 cm on each side. What is its volume?<br><br> Answer: 64cm^3

Anit [1.1K]

Answer:

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Explanation:

5 0

3 years ago

What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb

Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

$(\text{CH}_3)_3\text{N}$ in this question acts as a weak base. As seen in the equation in the question, $(\text{CH}_3)_3\text{N}$ produces $\text{OH}^{-}$ rather than $\text{H}^{+}$ when it dissolves in water. The concentration of $\text{OH}^{-}$ will likely be more useful than that of $\text{H}^{+}$ for the calculations here.

Finding the value of $[\text{OH}^{-}]$ from pH:

Assume that $\text{pK}_w = 14$ ,

$\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}$ .

$[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}$ .

Solve for $[(\text{CH}_3)_3\text{N}]_\text{initial}$ :

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}$

Note that water isn't part of this expression.

The value of Kb is quite small. The change in $(\text{CH}_3)_3\text{N}$ is nearly negligible once it dissolves. In other words,

$[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}$ .

Also, for each mole of $\text{OH}^{-}$ produced, one mole of $(\text{CH}_3)_3\text{NH}^{+}$ was also produced. The solution started with a small amount of either species. As a result,

$[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}$ .

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3}$ ,

$[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b}$ ,

$[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}$ .

8 0

3 years ago