At which temperature do particles stop moving entirely?

A 603.0 g/hr stream of liquid methyl alcohol, also called methanol, (CH3OH) at 8.10 atm and 26.0°C was held at constant pressure

mafiozo [28]

Answer:

$Q=\frac{1385270 J}{hr}$

Explanation:

Methanol's molecular weight: $M=32 g/mol$

Methanol's heat of vaporization: $\Delta H_{vap}=69.69 kJ/mol$

Ideal gas heat capacity (Cp):

$Cp=\frac{5}{2}*R$ where R is the gas constant

$Cp=\frac{5}{2}*8.314 \frac{J}{mol*C}=20.78\frac{J}{mol*C}$

The heat needed to vaporize and bring the gs to 210°C is:

$Q=\frac{603 g/hr}{32 g/mol}*(69690\frac{J}{mol}+20.78\frac{J}{mol*C}*(210-26)C)$

$Q=\frac{1385270 J}{hr}$

4 0

2 years ago

A certain drug has a half-life in the body of 3.5h. Suppose a patient takes one 200.Mg pill at :500PM and another identical pill

tekilochka [14]

Answer:

The amount of drug left in his body at 7:00 pm is 315.7 mg.

Explanation:

First, we need to find the amount of drug in the body at 90 min by using the exponential decay equation:

$N_{t} = N_{0}e^{-\lambda t}$

Where:

λ: is the decay constant = $ln(2)/t_{1/2}$

$t_{1/2}$ : is the half-life of the drug = 3.5 h

N(t): is the quantity of the drug at time t

N₀: is the initial quantity

After 90 min and before he takes the other 200 mg pill, we have:

$N_{t} = 200e^{-\frac{ln(2)}{3.5 h}*90 min*\frac{1 h}{60 min}} = 148.6 mg$

Now, at 7:00 pm we have:

$t = 7:00 pm - (5:00 pm + 90 min) = 30 min$

$N_{t} = (200 + 148.6)e^{-\frac{ln(2)}{3.5 h}*30 min*\frac{1 h}{60 min}} = 315.7 mg$

Therefore, the amount of drug left in his body at 7:00 pm is 315.7 mg (from an initial amount of 400 mg).

I hope it helps you!

3 0

2 years ago

Porfavor alguien que me explique los Cambios bioquimicos en los musculos ,sangre e higado durante los ejercicios fisicos y como

german

Answer:

hx see i own right to me I can send it it is not too good to be true that

Explanation:

iibhuuhhhucjdjdjrjnjxnxxhhhbbhhhhbhhbhhhet

8 0

3 years ago

At equilibrium, ________. At equilibrium, ________. all chemical reactions have ceased the rate constants of the forward and rev

balandron [24]

Answer:

<em>At equilibrium, the rate of the forward, and the reverse reactions are equal.</em>

Explanation:

In an equilibrium chemical reaction, the rate of forward reaction, is equal to the rate of reverse reaction. Note that the reactions does not cease at equilibrium, but rather, the reactants are converted to product, at the same rate at which the product is also being converted into the reactants in the reaction. When chemical equilibrium is reached, a careful calculation of the value of equilibrium constant is approximately equal to 1.

NB: If the value of equilibrium constant is far far greater than 1, then the reaction will favors more of the forward reaction, and if far far less than 1, the reaction will favor more of the reverse reaction.

6 0

2 years ago

The rate constant for the reaction 3A equals 4b is 6.00×10 how long will it take the concentration of a to drop from 0.75 to 0.2

Lelu [443]

This question is incomplete, the complete question is;

The rate constant for the reaction 3A equals 4B is 6.00 × 10⁻³ L.mol⁻¹min⁻¹.

how long will it take the concentration of A to drop from 0.75 to 0.25M ?

from the unit of the rate constant we know it is a second reaction order

OPTIONS

a) 2.2×10^−3 min

b) 5.5×10^−3 min

c) 180 min

d) 440 min

e) 5.0×10^2 min

Answer:

it will take 440 min for the concentration of A to drop from 0.75 to 0.25M

Option d) 440 min is the correct answer

Explanation:

Given that;

Rate constant K = 6.00 × 10⁻³ L.mol⁻¹min⁻¹

3A → 4B

given that it is a second reaction order;

k = 1/t [ 1/A - 1/A₀]

kt = [ 1/A - 1/A₀]

t = [ 1/A - 1/A₀] / k

K is the rate constant(6.00 × 10⁻³)

A₀ is initial concentration( 0.75 )

A is final concentration(0.25)

t is time required = ?

so we substitute our values into the equation

t = [ (1/0.25) - (1/0.75)] / (6.00 × 10⁻³)

t = 2.6666 / (6.00 × 10⁻³)

t = 444.34 ≈ 440 min {significant figures}

Therefore it will take 440 min for the concentration of A to drop from 0.75 to 0.25M

Option d) 440 min is the correct answer

6 0

3 years ago