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Mars2501 [29]
2 years ago
7

Draw the major product for the reaction of 1-butyne with water in the presence of catalytic TfOH (i.e., CF3SO3H). Then answer th

e additional question regarding this transformation.
Chemistry
1 answer:
Mademuasel [1]2 years ago
4 0

Answer:

2-Butanone

Explanation:

From the given information:

The presence of mercury as an acid catalyst brings about the addition of water to the triple bond which yields enol. Then, according to Markownikov's rule and after tautomerism has occurred, we have a methyl ketone ( 2- Butanone) as the product.

The answer regarding the transformation is addition and hydration.

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Freon-12, CF2Cl2, which has been widely used in air conditioning systems, is considered a threat to the ozone layer in the strat
Vilka [71]

Answer:

The root mean squared velocity for CF2Cl2 is  v_{rms}= 207.06 m/s

Explanation:

From the question we are told that

         The temperature is T = -65 ^oC = -65+273 = 208K

Root Mean Square velocity is mathematically represented as

      v _{rms} = \sqrt{\frac{3RT}{MW} }

 Where  T is the temperature

              MW is the molecular weight of gas

              R is the gas constant with a value of  R = 8.314 JK^{-1} mol^{-1}

For  CF2Cl2 its molecular weight is  0.121 kg/mol

     Substituting values

  v_{rms} = \sqrt{\frac{3 * 8.314 *208}{0.121} }

          v_{rms}= 207.06 m/s

5 0
2 years ago
Read 2 more answers
Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end point. If we assume that the acidit
Rasek [7]

Answer:

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

Explanation:

Volume of NaOH = 1.7 ml = 0.0017 L

Molarity of NaOH = 0.0811 M

Moles of NaOH = n

0.0811 M=\frac{n}{0.0017 L}

n = 0.0001378 mol

H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O

According to reaction, 2 mol of NaOH neutralize 1 mol of sulfuric acid.

Then 0.0001378 mol of NaOH will neutralize:

\frac{1}{2}\times 0.0001378 mol=6.8935\times 10^{-5} mol of sulfuric acid.

Concentration of sulfuric acid in the acid rain sample: x

x=\frac{6.8935\times 10^{-5}}{0.02 L}=0.0034467 mol/L

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

7 0
2 years ago
What is the mass of solute in 200.0 L of a 1.556-M solution of KBr
sergejj [24]
The molarity of KBr solution is 1.556 M
molarity is defined as the number of moles of solute in volume of 1 L solution.
the number of KBr moles in 1 L - 1.556 mol
Therefore in 200.0 L - 1.556 mol/L x 200.0 L = 311.2 mol
Molar mass of KBr - 119 g/mol 
mass of Kbr - 311.2 mol x 119 g/mol = 37 033 g
mass of solute therefore is 37.033 kg
4 0
2 years ago
How much energy is required to decompose 612 g of pcl3, according to the reaction below? the molar mass of pcl3 is 137.32 g/mol
Assoli18 [71]
The reaction is:

4 PCl3 (g) ---> P4(s) + 6 Cl2(g).

Now, you need to convert the mass of PCl3 into number of moles, for which you use the molar mass of PCl3 in this way:

number of moles = number of grams / molar mass =>

number of moles of PCl3 = 612 g / 137.32 g/mol = 4.4567 moles of PCl3.

Now use the proportion with the ΔH rxn given.

4 mol PCl3 / 1207 kJ = 4.4567 mol / x => x = 4.4567 mol * 1207 kJ / 4 mol = 1,344.8 kJ = 1.34 * 10^3 kJ.

Answer: 1.34 * 10 ^3 kJ (option d)
8 0
2 years ago
Read 2 more answers
A protein subunit from an enzyme is part of a research study and needs to be characterized. A total of 0.145 g of this subunit w
Hitman42 [59]

Answer:

The molar mass of the protein is 12982.8 g/mol.

Explanation:

The osmptic pressure is given by:

π=MRT

Where,

M: is molarity of the solution

R: the ideal gas constant (0.0821 L·atm/mol·K)

T: the temperature in kelvins

Hence, we look for molarity:

0.138 atm=M(0.0821\frac{l*atm}{mol*K} )(28+273K)

M=\frac{0.138atm}{(0.0821\frac{l*atm}{mol*K} )(301K)}= =5.584×10⁻³mol/l

As we have 2 ml of solution, we can get the moles quantity:

Moles of protein: 5.584×10⁻³\frac{mol}{l}\frac{1l}{1000ml}×2ml=1.117×10⁻⁵mol

Finally, the moles quantity is the division between the mass of the protein and the molar mass of the protein, so:

Moles=Mass/Molar mass

Molar mass= Mass/Moles=\frac{0.145g}{1.117*10^{-5}mol}=12982.8 g/mol

8 0
2 years ago
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