Ask question

Ask question

All categories

Ad libitum [116K]

3 years ago

9

For the following SN1 solvolysis reaction, select the correct rate equation and the overall reaction order (note: CH3OH is the s

olvent for the reaction). (CH3)3CCl + CH3OH → (CH3)3OCH3 + HCl

1 answer:

sineoko [7]3 years ago

8 0

Answer:

Explanation:

rate = K[(CH3)3CCl]

overall order is also first order

Since we are told that CH3OH is the solvent for the reaction, as such the rate law equation will only be written for (CH3)3CCl .

You might be interested in

What is 5L to mL ( chemistry )

zlopas [31]

There are 1000 mililiters in a liter, so 1000 ml for every liter, you have 5 liters, so:

5L*1000 = 5000 mL

5 0

3 years ago

Read 2 more answers

An ethylene glycol solution contains 21.4 g of ethylene glycol (C2H6O2) in 97.6 mL of water. (Assume a density of 1.00 g/mL for

8090 [49]

Answer: The freezing point and boiling point of the solution are $-6.6^0C$ and $101.8^0C$ respectively.

Explanation:

Depression in freezing point:

$T_f^0-T^f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_f$ = freezing point of solution = ?

$T^o_f$ = freezing point of water = $0^0C$

$k_f$ = freezing point constant of water = $1.86^0C/m$

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

$w_2$ = mass of solute (ethylene glycol) = 21.4 g

$w_1$ = mass of solvent (water) = $density\times volume=1.00g/ml\times 97.6ml=97.6g$

$M_2$ = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

$(0-T_f)^0C=1\times (1.86^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}$

$T_f=-6.6^0C$

Therefore,the freezing point of the solution is $-6.6^0C$

Elevation in boiling point :

$T_b-T^b^0=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_b$ = boiling point of solution = ?

$T^o_b$ = boiling point of water = $100^0C$

$k_b$ = boiling point constant of water = $0.52^0C/m$

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

$w_2$ = mass of solute (ethylene glycol) = 21.4 g

$w_1$ = mass of solvent (water) = $density\times volume=1.00g/ml\times 97.6ml=97.6g$

$M_2$ = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

$(T_b-100)^0C=1\times (0.52^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}$

$T_b=101.8^0C$

Thus the boiling point of the solution is $101.8^0C$

4 0

3 years ago

If the bright lines in the spectrum of oxygen are matched by lines in the dark-line spectrum of the sun, what does that indicate

lbvjy [14]

Hello young fellow friend I think the anwser is (C)

7 0

3 years ago

Read 2 more answers

A sample of propane (C3H8) has a mass of 0. 47 g. The sample is burned in a bomb calorimeter that has a mass of 1. 350 kg and a

Nady [450]

The amount of heat released by the sample has been 22.54 kJ. Thus, option C is correct.

The specific heat has been defined as the amount of heat required to raise the temperature of 1 gram of substance by 1 degree Celsius.

The specific heat has been expressed as:

$q=mc\Delta T$

<h3 /><h3>Computation for the heat absorbed</h3>

The iron and calorimeter are in side the closed system. Thus, the energy released by the sample, has been equivalent to the energy absorbed by the calorimeter.

$q_{released}=q_{absorbed}\\
q_{released}=m_{calorimeter}\;c_{calorimeter}\;\Delta T$

The given mass of calorimeter has been, $m_{calorimeter}=1350\;\rm g$

The specific heat of the calorimeter has been, $c_{calorimeter}=5.82\;\rm J/g^\circ C$

The change in temperature of the calorimeter has been, $\Delta T=2.87^\circ \rm C$

Substituting the values for heat released:

$q_{released}= 1350\;\text g\;\times\;5.82\;\text J/\text g^\circ \text C\;\times\;2.87^\circ \text C\\
q_{released}=22,549.5\;\text J\\
q_{released}}=22.54\;\rm kJ$

The amount of heat released by the sample has been 22.54 kJ. Thus, option C is correct.

Learn more about specific heat, here:

brainly.com/question/2094845

6 0

2 years ago

What is 1 major discovery in cosmology that relies on improvement in existing apparatus?

never [62]

The most dramatic astronomical development of the century thus far is the detection of gravitational waves from merging black holes at a distance of 400 Mpc, during the first science run of the advanced Laser Interferometer Gravitational-Wave Observatory.

The telescope was also very important. Galileo Galilei was the first person to use a telescope to look at celestial bodies (though he did not invent the telescope) and discovered the four brightest moons of Jupiter, proving that there are things in the Solar System that don't revolve around the Sun.

Physical cosmology is the branch of physics and astrophysics that deals with the study of the physical origins and evolution of the Universe. It also includes the study of the nature of the Universe on a large scale. In its earliest form, it was what is now known as "celestial mechanics", the study of the heavens.

Hope this helps you :)

3 0

3 years ago