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3 years ago

Which best describes the geologic time scale

Chemistry

2 answers:

pashok25 [27]3 years ago

5 0

Answer:

The correct answer about geologic time scale is that, it presents the correct sequence of events in the history of the earth.

Geologic Time scale is being recorded in the history of geologics. It is being broken into smaller and larger subdivisions and this helps us to know how events fit together.

In order to study the major events which are found in geologic history we should study the rock strata which is found around the world.

Explanation:

just olya [345]3 years ago

5 0

Answer:

the answer is B

Explanation:

The correct answer about geologic time scale is that, it presents the correct sequence of events in the history of the earth.

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Carbon tetrachloride reacts at high temperatures with oxygen to produce two toxic gases, phosgene and chlorine. CCl4(g) + (1/2)O

madam [21]

The question is incomplete, here is the complete question:

Carbon tetrachloride reacts at high temperatures with oxygen to produce two toxic gases, phosgene and chlorine.

$CCl_4(g)+\frac{1}{2}O_2(g)\rightleftharpoons COCl_2(g)+Cl_2(g);K_c=4.4\times 10^9$ at 1,000 K

Calculate Kc for the reaction $2CCl_4(g)+O_2(g)\rightleftharpoons 2COCl_2(g)+2Cl_2(g)$

<u>Answer:</u> The value of $K_c'$ for the final reaction is $1.936\times 10^{19}$

<u>Explanation:</u>

The given chemical equations follows:

$CCl_4(g)+\frac{1}{2}O_2(g)\rightleftharpoons COCl_2(g)+Cl_2(g);K_c$

We need to calculate the equilibrium constant for the equation, which is:

$2CCl_4(g)+O_2(g)\rightleftharpoons 2COCl_2(g)+2Cl_2(g)$

As, the final reaction is the twice of the initial equation. So, the equilibrium constant for the final reaction will be the square of the initial equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_c'=(K_c)^2$

We are given:

$K_c=4.4\times 10^9$

Putting values in above equation, we get:

$K_c'=(4.4\times 10^9)^2=1.936\times 10^{19}$

Hence, the value of $K_c'$ for the final reaction is $1.936\times 10^{19}$

3 0

3 years ago

Ill give u brainliest and all my points! pls help

dexar [7]

http://www.atnf.csiro.au/outreach/education/senior/cosmicengine/stars_hrdiagram.html. This will give u the answer, just go to link

6 0

3 years ago

A distance of one centimeter is the same as ?

rodikova [14]

C because its meters right

4 0

3 years ago

Determine the pH of the resulting solution if 25 mL of 0.400 M strychnine (C21H22N2O2) is added to 50 mL of 0.200 M HCl? Assume

DIA [1.3K]

Answer:

pH = 4.56

Explanation:

The strychnine reacts with HCl as follows:

C₂₁H₂₂N₂O₂ + HCl ⇄ C₂₁H₂₂N₂O₂H⁺ + Cl⁻

<em />

For strychnine buffer:

pOH = 5.74 + log [C₂₁H₂₂N₂O₂H⁺] / [C₂₁H₂₂N₂O₂]

Initial moles of C₂₁H₂₂N₂O₂ are:

0.025L * (0.400 mol / L) = 0.01 moles C₂₁H₂₂N₂O₂

And of HCl are:

0.05L * (0.200 mol / L) = 0.01 moles HCl

That means after the reaction, you will have just 0.01 moles of C₂₁H₂₂N₂O₂H⁺ in 50mL + 25mL = 0.075L. And molarity is:

[C₂₁H₂₂N₂O₂H⁺] = 0.01 mol / 0.075L = 0.1333M

This conjugate acid, is in equilibrium with water as follows:

C₂₁H₂₂N₂O₂H⁺(aq) + H₂O(l) ⇄ C₂₁H₂₂N₂O₂ + H₃O⁺

<em />

<em>Where Ka = Kw / Kb = 1x10⁻¹⁴ / 1.8x10⁻⁶ = 5.556x10⁻⁹</em>

<em />

Ka is defined as:

Ka = 5.556x10⁻⁹ = [C₂₁H₂₂N₂O₂] [H₃O⁺] / [C₂₁H₂₂N₂O₂H⁺]

In equilibrium, concentrations are:

C₂₁H₂₂N₂O₂ = X

H₃O⁺ = X

C₂₁H₂₂N₂O₂H⁺ = 0.1333M - X

Replacing in Ka expression:

5.556x10⁻⁹ = [X] [X] / [0.1333M - X]

7.39x10⁻¹⁰ - 5.556x10⁻⁹X = X²

7.39x10⁻¹⁰ - 5.556x10⁻⁹X - X² = 0

Solving for X:

X = - 2.72x10⁻⁵M → False solution. There is no negative concentrations

X = 2.72x10⁻⁵M → Right solution.

As H₃O⁺ = X

H₃O⁺ = 2.72x10⁻⁵M

And pH = -log H₃O⁺

4 0

3 years ago

Someone has left a shiny metal bowl outside in the sun. Which may have the potential to damage your eyes: looking at the outside

STatiana [176]

Answer:

Looking at the inside of the bowl

Explanation:

The curvature of the bowl would likely focus the light to a single point or area above it. This higher intensity area would be more dangerous to look at.

3 0

3 years ago

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