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3 years ago

Draw one product that you would expect from the reaction of 1 mol of 1,3-butadiene and 1 mol of h2o, cat. h2so4?

1 answer:

3 0

1,3-butadiene is the simplest conjugated diene and undergoes 1,4 addition reaction in acidic environment.
Chemical reaction: CH₂=CH-CH=CH₂ + H₂O → CH₃-CH=CH-CH₂-OH.
CH₂=CH-CH=CH₂ - 1,3-butadiene.
CH₃-CH=CH-CH₂-OH - 2-buten-1-ol.
Diene or diolefin is a hydrocarbon that has two carbon double bonds.

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What energy does a rock resting on a hilltop store?

Hunter-Best [27]

Answer:

gravitational energy

Explanation:

It is gravitational (potential) energy because of the place that the rock holds in the gravitational field. It has potential to move downward, because of gravity.

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3 years ago

Calculate the molarity of a solution with 233.772g sodium chloride dissolved in 2,000mL of water

LekaFEV [45]

Answer:

The molarity is 2M

Explanation:

First , we calculate the weight of 1 mol of NaCl:

Weight 1mol NaCl= Weight Na + Weight Cl= 23 g+ 35, 5 g= 58, 5 g/mol

58,5 g---1 mol NaCl

233,772 g--------x= (233,772 g x1 mol NaCl)/58,5 g= 4 mol NaCl

A solution molar--> moles of solute in 1 L of solution:

2 L-----4 mol NaCl

1L----x0( 1L x4mol NaCl)/4L =2moles NaCl---> 2 M

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3 years ago

Calculate the mass, in grams, of Ag2CrO4 that will precipitate when 50.0mL of 0.20M AgNO3 solution is mixed with 40.0mL of 0.10M

Darina [25.2K]

Answer:

1.327 g Ag₂CrO₄

Explanation:

The reaction that takes place is:

2AgNO₃(aq) + K₂CrO₄(aq) → Ag₂CrO₄(s) + 2KNO₃(aq)

First we need to identify the limiting reactant:

We have:

0.20 M * 50.0 mL = 10 mmol of AgNO₃

0.10 M * 40.0 mL = 4 mmol of K₂CrO₄

If 4 mmol of K₂CrO₄ were to react completely, it would require (4*2) 8 mmol of AgNO₃. There's more than 8 mmol of AgNO₃ so AgNO₃ is the excess reactant. That makes K₂CrO₄ the limiting reactant.

Now we calculate the mass of Ag₂CrO₄ formed, using the limiting reactant:

4 mmol K₂CrO₄ * $\frac{1mmolAg_2CrO_4}{1mmolK_2CrO_4} *\frac{331.73mg}{1mmolAg_2CrO_4}$ = 1326.92 mg Ag₂CrO₄

1326.92 mg / 1000 = 1.327 g Ag₂CrO₄

7 0

2 years ago

قدري بلا لون الجزء الثالث الحلقه ٩

nevsk [136]

Answer: Arabic

My destiny without color, the third part 9 episode

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3 years ago

Determine the volume of hydrogen collected at STP and SATP, and the molar volume at STP and SATP

mario62 [17]

The other given data are:
Room temperature (°C) 22.0; pressure (kPa) 100.5; Water vapor pressure at 22 °C (2.60 kPa); Mass of Mg ribbon (0.05g); Volume of hydrogen gas (mL) 48.3

Vstp = 22.4 L/mol, and Vsatp = 24.8 L/mol Vstp:0 degrees Celsius and 101.325 kPa while Vsatp: 25 degrees Celsius and 100 kPa. 100.5 kPa x 760 mm Hg/101.325 kP = 753.8119911 mmHg

2HCl(g) + Mg(s) -> H2(g) + MgCl2(aq) nMg = 0.0020571899 mol nH2 = 0.0020571899 mol Wet H2 pressure = 100.5 kPa Dry H2 pressure = 97.9 kPa

So to determine the volume of hydrogen gas collected at STP: V2 = (753.8119911 mmHg)(48.3 mL)(273 K) / (295 K)(760 mmHg) = 44.33403003 mL.

Molar volume at STP: mol/L = 44.33403003 mL / 0.0020571899 mol
= 21550.77176 mL/mol
= 21.55077176 L/mol
To determine volume of hydrogen gas collected at SATP:
V2 = (100.5 kPa)(48.3 mL)(298 K) / (295K)(100kPa) = 49.03514237 mL

Molar volume at SATP: mol/L = 49.03514237 mL / 0.0020571899 mol
= 23835.98246 mL/mol
= 23.8359824 L/mol

4 0

3 years ago