What is the mass number of the isotope lithium -7?

The rate of decay of a chemical involved in a reaction that is second order (bimolecular) in one reactant A is given by: = k [AJ

anyanavicka [17]

Explanation:

$2A\rightarrow products$

According to mass action,

$\textrm{rate}=-\dfrac{\Delta[\textrm A]}{2\Delta t}=k[\textrm A]^2$

Where, k is the rate constant

So,

$\dfrac{d[A]}{dt}=-k[A]^2$

Integrating and applying limits,

$\int_{[A_t]}^{[A_0]}\frac{d[A]}{[A]^2}=-\int_{0}^{t}kdt$

we get:

$\dfrac{1}{[A]} = \dfrac{1}{[A]_0}+kt$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Half life is the time when the concentration reduced to half.

So, $[A_t]=\frac{1}{2}\times [A_0]$

Applying in the equation as:

$t_{1/2}=\dfrac{1}{k[A_o]}$

7 0

4 years ago

Now imagine that you're waiting for your soup to cool down, and you leave it on the counter while you go do your homework. Your

Vlad [161]

Explanation:

Generally, heat flows from a hot environment to a cold (lesser temperature) environment. In this case, the soup is the hot environment and the air is the cold temperature.

Heat would continue to flow from one environment to another until thermal equilibrium is reached. At this thermal equilibrium, both environments would have the same temperature.

4 0

3 years ago

What is the final volume (L) of a 1.00 L system at 315 K and 1.10 atm if STP conditions are established?

LenaWriter [7]

Answer:

0.95L

Explanation:

Data obtained from the question include:

V1 (initial volume) = 1L

T1 (initial temperature) = 315K

P1 (initial pressure) = 1.10 atm

T2 (final temperature) = stp = 273K

P2 (final pressure) = stp = 1atm

V2 (final volume) =?

Using the general gas equation P1V1/T1 = P2V2/T2, the final volume of the system can be obtained as follow:

P1V1/T1 = P2V2/T2

1.1 x 1/315 = 1 x V2/273

Cross multiply to express in linear form.

315 x V2 = 1.1 x 273

Divide both side by 315

V2 = (1.1 x 273) /315

V2 = 0.95L

Therefore, the final volume of the system if STP conditions are established is 0.95L

5 0

3 years ago

(LO 3N, 4G, 4O) Aluminum sulfate is also involved in dying fabrics. The gelatinous precipitate formed in the reaction with dilut

brilliants [131]

9.4 × 10⁻³ mg (0.73 mmoles) of Al(OH)₃ is formed

Explanation:

We have the following chemical reaction:

Al₂(SO₄)₃(aq) + 6 NaOH(aq) → 2 Al(OH)₃(s) + 3 Na₂SO₄(aq)

The precipitate mentioned by the problem is aluminium hydroxide Al(OH)₃.

Now to determine the number of moles of sodium hydroxide NaOH we use the following formula:

molar concentration = number of moles / volume

number of moles = molar concentration × volume

number of moles of NaOH = 0.088 M × 25 mL = 2.2 mmoles

number of moles of Al₂(SO₄)₃ = 5.6 × 10⁻³ moles = 5.6 mmoles (found in the problem text)

We see from the chemical reaction that 1 mole of Al₂(SO₄)₃ requires 6 moles of NaOH so 5.6 mmoles of Al₂(SO₄)₃ would require 6 times more NaOH which is 33.6 mmoles and we have only 2.2 mmoles. The limiting reactant will be NaOH.

Now we devise the following reasoning:

if 6 mmoles of NaOH produces 2 mmoles of Al(OH)₃

then 2.2 mmoles of NaOH produces X mmoles of Al(OH)₃

X = (2.2 × 2) / 6 = 0.73 mmoles of Al(OH)₃

mass of Al(OH)₃ = number of moles / molecular weight

mass of Al(OH)₃ = 0.73 / 78

mass of Al(OH)₃ = 9.4 × 10⁻³ mg

Learn more:

precipitation reaction

brainly.com/question/10400269

7 0

3 years ago

A student is doing an experiment to determine the effects of temperature on an object. He writes down that the initial temperatu

bogdanovich [222]

Answer:

1) The Kelvin temperature cannot be negative

2) The Kelvin degree is written as K, not ºK

Explanation:

The temperature of an object can be written using different temperature scales.

The two most important scales are:

- Celsius scale: the Celsius degree is indicated with ºC. It is based on the freezing point of water (placed at 0ºC) and the boiling point of water (100ºC).

- Kelvin scale: the Kelvin is indicated with K. it is based on the concept of "absolute zero" temperature, which is the temperature at which matter stops moving, and it is placed at zero Kelvin (0 K), so this scale cannot have negative temperatures, since 0 K is the lowest possible temperature.

The expression to convert from Celsius degrees to Kelvin is:

$T(K)=T(^{\circ}C)+273.15$

Therefore in this problem, since the student reported a temperature of -3.5 ºK, the errors done are:

1) The Kelvin temperature cannot be negative

2) The Kelvin degree is written as K, not ºK

6 0

4 years ago