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2 years ago

Explain why c6h5ch2ch2br is not formed during the radical bromination of c6h5ch2ch3. select the single best answer.

1 answer:

7 0

Hi!

The radical bromination reaction of C₆H₅CH₂CH₃ is performed through a mechanism in which radical reactions are involved. This compound is an alkylbenzene compound, and the carbon that is more reactive towards radical bromination is the carbon bonded to the aromatic ring because in the reaction mechanism the intermediaries are stabilized by resonance in the aromatic ring.

A terminal substitution will not occur because substitution there will not be stabilized by resonance. The compound that will be formed in this reaction would be:

C₆H₅CH₂CH₃ + Br₂ → C₆H₅CH₂(Br)CH₃ + HBr

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Which of the following compounds would be most effective in lowering the melting point of ice on roads? a) CaCl2 b) NaCl C) K3PO

Mrac [35]

I'm not 100% sure on this, but I would go with C) NaCl.
NaCl is a salt, and that is used to melt the ice on the roads. Hope this helps!

7 0

2 years ago

Explain one way the water cycle affects climate. Use complete sentences.

ladessa [460]

Answer:

The water cycle will affect the climate by causing rain and snow.

Explanation:

if tempatures increase more rain fall will occur because more water evaporates in the air.

4 0

1 year ago

The molecular weight of a gas is ________ g/mol if 6.7 g of the gas occupies 6.3 l at stp.

PIT_PIT [208]

At STP, also known as standard temperature and pressure, 1 mole of a gas occupies 22.4 L. Since we are given with the volume of 6.3L, we calculate the amount of gas in mol.
n = (6.3L)/ (22.4L/mol) = 0.28125 mol
We are given with the mass of 6.7 g. Therefore, the molar mass or molecular weight of the gas is equal to,
6.7g/0.28125 mol = 23.82 g/mol

6 0

2 years ago

Anyway, which is a spectator ion involved in the reaction of k2cro4(aq) and ba(no3)2(aq)?

marissa [1.9K]

Nitrate ions $\text{NO}_3^{-}$ and potassium ions $\text{K}^{+}$ .

<h3>Explanation</h3>

Potassium dichromate undergoes a double-decomposition reaction with barium nitrate to produce barium dichromate and potassium nitrate. The reaction is possible due to the low solubility of barium dichromate that precipitate out of the solution readily after its production.

Start with the balanced chemical equation for this process:

$\text{K}_2\text{CrO}_4 \; (aq) + \text{Ba}(\text{NO}_3)_2 \; (aq) \to \text{Ba}(\text{CrO}_4)_2 \; (s) + 2 \; \text{KNO}_3 \; (aq)$

Rewrite the chemical equation as an ionic one; express all soluble salts- those with state symbol (aq)- as their constituting ions while leaving the insoluble (s) intact.

$2 \; \text{K}^{+}\; (aq)+ 2\; \text{CrO}_4^{-} \; (aq) + \text{Ba}^{2+} \; (aq) + 2 \; \text{NO}_3^{-} \; (aq) \\ \to \text{Ba}(\text{CrO}_4)_2 \; (s) + 2 \; \text{K}^{+} \; (aq) + 2 \; \text{NO}_3^{-} \; (aq)$

$\text{K}^{+}$ and $\text{CrO}_4^{-}$ are found on both sides of the equation by the same quantity. The two ions thus took no part in the net reaction and act as spectator ions.

5 0

2 years ago

A mixture of CH4 and H2O is passed over a nickel catalyst at 1000 K. The emerging gas is collected in a 5.00L flask and is found

Harman [31]

Answer:

Kc =<u> 3.74*10⁻³ </u>

Kp = 25.21

Explanation:

Step 1: Data given

Temperature = 1000 K

Volume = 5.00 L

Mass of CO = 8.62 grams

Mass of H2 = 2.60 grams

Mass of CH4 = 43.0 grams

Mass of H2O = 48.4 grams

Kc = [CO]*[H₂]³ / ([CH₄]∙*H₂O])

Kp = p(CO)*p(H₂)³ / (p(CH₄)*p(H₂O) )

Step 2: The balanced equation

CH₄ + H₂O ⇄ CO + 3 H₂

Step 3: Calculate number of moles

The number of moles of each compund in the equilibrium mixture are:

Moles = mass / molar mass

n(CH₄) = 43.0g / 16g/mol = 2.688mol

n(H₂O) = 48.4g / 18g/mol = 2.689mol

n(CO) = 8.62g/28g/mol = 0.308mol

n(H₂) = 2.60g / 2g/mol = 1.3mol

Step 4: Calculate concentrations at equilibrium

So the equilibrium concentrations are:

Concentration = moles / volume

[CH₄] = 2.688mol/5L = 0.5376 M

[H₂O] = 2.689mol/5L = 0.5378M

[CO] = 0.308mol/5L = 0.0616M

[H₂) = 1.3mol/5L = 0.26M

Step 5: Calculate Kc

Kc = 0.0616 ∙ (0.26)³ / (0.5376∙0.5378) = <u>3.74*10⁻³ </u>

Step 5: Calculate partial pressure

Partial pressures in equilibrium can be found from ideal gas law:

p(X) = n(X)∙R∙T/V = [X]∙R∙T

=> p(CH₄) = [CH₄]∙R∙T = 0.5376mol/L * 0.082 06Latm/molK ∙ 1000K = 44.11 atm

p(H₂O) = [H₂O]∙R∙T = 0.5738mol/L * 0.082 06Latm/molK * 1000K = 44.13 atm

p(CO) = [CO]∙R∙T = 0.0616mol/L * 0.082 06Latm/molK * 1000K = 5.05atm

p(H₂) = [CO]∙R∙T = 0.26mol/L * 0.082 06Latm/molK * 1000K = 21.34atm

Step 5: Calculate Kp

Kp = p(CO)*p(H₂)³ / (p(CH₄)*p(H₂O) )

Kp = 5.05*21.34³ / (44.11*44.13 ) = 25.21

8 0

2 years ago