S308<br> what’s the name to this

"Alkenes" ________.are reasonably soluble in water are relatively polar compounds have lower boiling points than alcohols of sim

aleksandr82 [10.1K]

Answer:

The answer to your question is: the third option is correct.

Explanation:

"Alkenes" ________.

are reasonably soluble in water This option is incorrect, alkenes are insoluble in water.

are relatively polar compounds This option is wrong, alkenes are not polar molecules.

have lower boiling points than alcohols of similar molecular weight This option is true, alkenes have a lower boiling point than alcohols.

both have lower boiling points than alcohols of similar molecular weight and are relatively polar compounds. This option is incorrect because is combining true with false facts.

8 0

3 years ago

For the following reaction, 4.64 grams of oxygen gas are mixed with excess benzene (C6H6). The reaction yields 3.95 grams of car

Reil [10]

Answer:

Theoretical yield for CO₂ is 5.10g

Explanation:

Reaction: 2C₆H₆(l) + 15O₂(g) → 12CO₂(g) + 6H₂O(g)

We convert the mass of oxygen to moles:

4.64 g /32 g/mol = 0.145 moles of O₂

Let's find out the 100% yield reaction of CO₂ (theoretical yield)

Ratio is 15:12. So let's make this rule of three:

15 moles of O₂ can produce 12 moles of CO₂

Therefore 0.145 moles of oxygen will produce (0.145 . 12) /15 = 0.116 moles

We convert the moles to mass: 0.116 mol . 44 g / 1mol = 5.10 g

6 0

2 years ago

Read 2 more answers

A second- order reaction of the type A + B -->P was carried out in a solution that was initially 0.075 mol dm^-3 in A and 0.0

andriy [413]

Answer:

a) 16.2 dm^3/mol*h

b) 6.1 × 10^3 s, 2.5 × 10^3 s (it is different to the hint)

Explanation:

We can use the integrated rate equation in order to obtain k.

For the reaction A+ B --> P the reaction rate is written as

Rate = $-\frac{dC_A}{dt} = -\frac{dC_B}{dt} = \frac{dC_P}{dt} = kC_AC_B$

If $C_{A0}$ and $C_{B0}$ are the inital concentrations and x the concentration reacted at time t, so $C_A=C_{A0} -x$ and $C_B=C_{B0} -x$ and the rate at time t is written as:

$-\frac{dx}{dt} =-k(C_{A0} -x)(C_{B0}-x)$

Separating variables and integrating

$\int\limits^x_0 {\frac{1}{(C_{A0}-x)(C_{B0}-x)} } \, dx = \int\limits^t_0 {k} \, dt$

The integral in left side is solved by partial fractions, it can be used integral tables

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}}{C_{A0}-x}-ln\frac{C_{B0}}{C_{B0}-x}) =kt$

Using logarithm properties (ln x - ln y = ln(x/y))

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}C_{B}}{C_{A}C_{B0}}) =kt$

Using the given values k can be calculated. But the data seems inconsistent since if the concentration of A changes from 0.075 to 0.02 mol dm^-3 it implies that 0.055 mol dm^-3 of A have reacted after 1 h, so according to the reaction given the same quantity of B should react, and we only have a $C_{B0}$ of 0.05 mol dm^-3.

Assuming that the concentration of B fall to 0.02 mol dm^-3 (and not the concentration the A). So we arrive to the answer given in the Hint.

So, the values given are t= 1, $C_{A0}=0.075$ , $C_{B0}=0.05$ , $C_{B}=0.02$ , it implies that the quantity reacted, x, is 0.03 and $C_{A}=0.075-x = 0.045$ . Then, the value of k would be

$kt = \frac{1}{0.05-0.075}(ln\frac{0.075*0.02}{0.045*0.05})$

$k = 16.21 \frac{dm^3}{mol*1h}$

b) the question b requires calculate the time when the concentration of the specie is half of the initial concentration.

For reactant A, It is solved with the same equation

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}C_{B}}{C_{A}C_{B0}}) =kt$

but suppossing that $C_A= C_{A0}/2=0.0375$ so $C_B=C_{B0}- C_{A0}/2=0.0125$ , k=16.2 and the same initial concentrations. Replacing in the equation

$t=\frac{1}{16.2(0.05-0.075)}(ln\frac{0.075*0.0125}{0.0375*0.05})$

$t=1.71 h = 1.71*3600 s = 6.1*10^3 s$

For reactant B, $C_B= C_{B0}/2=0.025$ so $C_A=C_{A0}- C_{B0}/2=0.05$ , k=16.2 and the same initial concentrations. Replacing in the equation

$t=\frac{1}{16.2(0.05-0.075)}(ln\frac{0.075*0.025}{0.05*0.05})$

$t=0.71 h = 0.7*3600 s = 2.5*10^3 s$

Note: The procedure presented is correct, despite of the answer be something different to the given in the hint, I obtain that result if the k is 19.2... (maybe an error in calculation of given numbers)

3 0

2 years ago

A 0.239 gram sample of a gas in a 100-mL flask exerts a pressure of 603 mm Hg at 14 degrees Celsius. What is the gas?

mamaluj [8]

Answer:

Br

Explanation:

Given data:

Mass of gas = 0.239 g

Volume of gas = 100 mL

Pressure exerted by gas = 603 mmHg

Temperature of gas = 14 °C

What is gas = ?

Solution:

Volume of gas = 100 mL (100mL ×1 L/1000 mL= 0.1 L)

Pressure exerted by gas = 603 mmHg (603/760 = 0.79 atm)

Temperature of gas = 14°C ( 14+273 = 287 K)

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K

T = temperature in kelvin

now we will calculate the number of moles.

n = PV/RT

n = 0.79 atm × 0.1 L / 0.0821 atm.L/ mol.K × 287 K

n = 0.079 /23.563 /mol

n = 0.003 mol

Molar mass of gas:

Number of moles = mass/molar mass

0.003 mol = 0.239 g/ molar mass

Molar mass = 0.239 g/ 0.003 mol

Molar mass = 79.7 g/mol

The molar mas of Br is 79.9 g/mol so it is closer to 79.7 thus given gas is Br.

6 0

2 years ago

David’s mom wants to calculate how much it will cost to drive from Los Angeles, CA, to San Francisco, CA. Gas costs $4. 00 a gal

bixtya [17]

The additional information required to calculate the cost of the drive has been distance between the two cities. Thus, option B is correct.

The distance traveled by the object set the cost per unit has been given by the product of the distance and the cost.

<h3>Detail required for calculation cost</h3>

The given question has known cost of the fuel per gallon.

The mileage of the car has been known.

The distance traveled by the car has not been known. Thus, to calculate the cost of the drive, the distance between the two points has to be known. Thus, option B is correct.

Learn more about distance traveled, here:

brainly.com/question/1214027

3 0

2 years ago

S308 what’s the name to this