Which group on the periodic table has the most reactive metals?

A gas occupies a volume of 2.4 L at 14.1 kPa. What volume will the gas occupy at 84.6 kPa?

Naddik [55]

Answer:

0.40 L

Explanation:

Boyle's law for gases states that, at constant temperature, the volume and pressure of a fixed amount of gas are inversely related.

Mathematically, that is:

PV = constant

P₁V₁ = P₂V₂

Here, you have:

V₁ = 2.4 L

P₁ = 14.1 Kpa

P₂ = 84.6 KPa

V₂ = ?

Then, you can solve for V₂:

V₂ = P₁V₁ / P₂

Substitute and compute:

V₂ = 14.1 KPa × 2.4L / 84.6 KPa = 0.40 L ← answer

3 0

3 years ago

COCl2(g) decomposes according to the equation above. When pure COCl2(g) is injected into a rigid, previously evacuated flask at

mestny [16]

Answer: The value of $K_p$ for the reaction at 690 K is 0.05

Explanation:

We are given:

Initial pressure of $COCl_2$ = 1.0 atm

Total pressure at equilibrium = 1.2 atm

The chemical equation for the decomposition of phosgene follows:

$COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)$

Initial: 1 - -

At eqllm: 1-x x x

We are given:

Total pressure at equilibrium = [(1 - x) + x+ x]

So, the equation becomes:

$[(1 - x) + x+ x]=1.2\\\\x=0.2atm$

The expression for $K_p$ for above equation follows:

$K_p=\frac{p_{CO}\times p_{Cl_2}}{p_{COCl_2}}$

$p_{CO}=0.2atm\\p_{Cl_2}=0.2atm\\p_{COCl_2}=(1-0.2)=0.8atm$

Putting values in above equation, we get:

$K_p=\frac{0.2\times 0.2}{0.8}\\\\K_p=0.05$

Hence, the value of $K_p$ for the reaction at 690 K is 0.05

3 0

3 years ago

Why copper should not be disposed of in landfill sites?

Dafna1 [17]

Copper is a heavy metal that can harm many people. This process of heavy metals building in your body is called bioaccumulation and the process of this building in the food chain is called biomagnification. Its hard to rid of this heavy metal because it is very dense, therefore, cannot breakdown in our bodies. Heavy metals get consumed by producers and then reach to the top of the pyramid (us) so we have the highest risk of getting diseases and so on. Make sure you explain that heavy metals get passed from industrial purposes where companies do not dispose of waste properly.
Source: Exam as well
Good luck!

4 0

3 years ago

Read 2 more answers

For the reaction 2N2O5(g) <---> 4NO2(g) + O2(g), the following data were colected:

KonstantinChe [14]

Answer:

a) The reaction is first order, that is, order 1. Option C is correct.

b) The half life of the reaction is 23 minutes. Option B is correct

c) The initial rate of production of NO2 for this reaction is approximately = (3.7 × 10⁻⁴) M/min. Option has been cut off.

Explanation:

First of, we try to obtain the order of the reaction from the data provided.

t (minutes) [N2O5] (mol/L)

0 1.24x10-2

10 0.92x10-2

20 0.68x10-2

30 0.50x10-2

40 0.37x10-2

50 0.28x10-2

70 0.15x10-2

Using a trial and error mode, we try to obtain the order of the reaction. But let's define some terms.

C₀ = Initial concentration of the reactant

C = concentration of the reactant at any time.

k = rate constant

t = time since the reaction started

T(1/2) = half life

We Start from the first guess of zero order.

For a zero order reaction, the general equation is

C₀ - C = kt

k = (C₀ - C)/t

If the reaction is indeed a zero order reaction, the value of k we will obtain will be the same all through the set of data provided.

C₀ = 0.0124 M

At t = 10 minutes, C = 0.0092 M

k = (0.0124 - 0.0092)/10 = 0.00032 M/min

At t = 20 minutes, C = 0.0068 M

k = (0.0124 - 0.0068)/20 = 0.00028 M/min

At t = 30 minutes, C = 0.0050 M

k = (0.0124 - 0.005)/30 = 0.00024 M/min

It's evident the value of k isn't the same for the first 3 trials, hence, the reaction isn't a zero order reaction.

We try first order next, for first order reaction

In (C₀/C) = kt

k = [In (C₀/C)]/t

C₀ = 0.0124 M

At t = 10 minutes, C = 0.0092 M

k = [In (0.0124/0.0092)]/10 = 0.0298 /min

At t = 20 minutes, C = 0.0068 M

k = 0.030 /min

At t = 30 minutes, C = 0.0050 M

k = 0.0303

At t = 40 minutes

k = 0.0302 /min

At t = 50 minutes,

k = 0.0298 /min

At t = 60 minutes,

k = 0.031 /min

This shows that the reaction is indeed first order because all the answers obtained hover around the same value.

The rate constant to be taken will be the average of them all.

Average k = 0.0302 /min.

b) The half life of a first order reaction is related to the rate constant through this relation

T(1/2) = (In 2)/k

T(1/2) = (In 2)/0.0302

T(1/2) = 22.95 minutes = 23 minutes.

c) The initial rate of production of the product at the start of the reaction

Rate = kC (first order)

At the start of the reaction C = C₀ = 0.0124M and k = 0.0302 /min

Rate = 0.0302 × 0.0124 = 0.000374 M/min = (3.74 × 10⁻⁴) M/min

3 0

3 years ago

Rank the following from lowest to highest anticipated boiling point: C8H18, CH, C16H34,CH4,C4H10

viva [34]

The relative strength of intermolecular forces such as ionic, hydrogen bonding, dipole-dipole interaction and Vander Waals dispersion force affects the boiling point of a compound. For this case, the longer the chain the higher the boiling point.

CH, CH4, C4H10, C8H18, C16H34

Hope this answers the question. Have a nice day.

8 0

3 years ago