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3 years ago

13

a _ is a possible explanation for a problem using what you know and what you observe a-conclusion b-thesis c-theory d- hypothesi

s

1 answer:

Andrej [43]3 years ago

7 0

I'm pretty sure its A

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(a) ions in a certain volume of .20M NaCl (aq) are represented in the box above on the left. in the box above on the right, draw

mamaluj [8]

Answer:

The first picture attached is the diagram that accompanies the question.

The<u> second picture attached</u> is the diagram with the answer.

Explanation:

In the box on the left there are 8 Cl⁻ ions and 8 Na⁺ ions.

The dissociaton equation for NaCl(aq) is:

NaCl (aq) → Na⁺ (aq) + Cl⁻(aq)

The dissociation equation for CaCl₂ (aq) is:

CaCl₂ (aq) → Ca²⁺ (aq) + 2Cl⁻(aq)

A 0.10MCaCl₂ (aq) solution will have half the number of CaCl₂ units as the number of NaCl units in a 0.20M NaCl (aq) solution.

Thus, while the 0.20M NaCl (aq) solution yields 8 ions of Na⁺ and 8 ions of Cl⁻, the 0.10MCaCl₂ (aq) solution will yield 4 ions of Ca²⁺ (half because the concentration if half) and 8 ions of Cl⁻ (first take half and then multiply by 2 because the dissociation reaction).

Thus, your drawing must show 4 dots representing Ca²⁺ ions and 8 dots representing Cl⁻ ions in the box on the right.

4 0

3 years ago

The rate constant for the second-order reaction: 2NOBr(g) → 2NO(g) + Br2(g) is 0.80/(M · s) at 10°C. Starting with a concentrati

a_sh-v [17]

Answer : The concentration of NOBr after 95 s is, 0.013 M

Explanation :

The integrated rate law equation for second order reaction follows:

$k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)$

where,

k = rate constant = $0.80M^{-1}s^{-1}$

t = time taken = 95 s

[A] = concentration of substance after time 't' = ?

$[A]_o$ = Initial concentration = 0.86 M

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

$0.80=\frac{1}{95}\left (\frac{1}{[A]}-\frac{1}{(0.86)}\right)$

[A] = 0.013 M

Hence, the concentration of NOBr after 95 s is, 0.013 M

4 0

3 years ago

Malia was able to make a paperclip float on the surface of water. What will most likely happen to the paperclip if a drop of dis

Arturiano [62]

Answer:

Its A

Explanation:

I just took the test

5 0

3 years ago

Explain the process of fossil fuel formation.

topjm [15]

Explanation:

Fossil fuel is an overall term for covered ignitable geologic stores of natural materials, framed from rotted plants and creatures that have been changed over to unrefined petroleum, coal, flammable gas, or weighty oils by introduction to warmth and weight in the world's outside more than a huge number of years.

The consuming of petroleum products by people is the biggest wellspring of emanations of carbon dioxide, which is one of the ozone depleting substances that permits radiative compelling and adds to an unnatural weather change.

A little bit of hydrocarbon-based powers are biofuels gotten from climatic carbon dioxide, and consequently don't build the net measure of carbon dioxide in the environment.

6 0

3 years ago

Phosgene (COCl2) is a toxic substance that forms readily from carbon monoxide and chlorine at elevated temperatures: CO(g) + Cl2

lapo4ka [179]

Answer: Concentration of $CO$ = 0.328 M

Concentration of $Cl_2$ = 0.328 M

Concentration of $COCl_2$ = 0.532 M

Explanation:

Moles of $CO$ and $Cl_2$ = 0.430 mole

Volume of solution = 0.500 L

Initial concentration of $CO$ and $Cl_2$ = $\frac{moles}{volume}=\frac{0.430}{0.500}=0.860M$

The given balanced equilibrium reaction is,

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

Initial conc. 0.860M 0.860M 0

At eqm. conc. (0.860-x) M (0.860-x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[COCl_2]}{[CO][Cl_2]}$

Now put all the given values in this expression, we get :

$4.95=\frac{x}{(0.860-x)^2}$

By solving the term 'x', we get :

x = 0.532 M

Thus, the concentrations of $CO,Cl_2\text{ and }COCl_2$ at equilibrium are :

Concentration of $CO$ = (0.860-x) M =(0.860-0.532) M = 0.328 M

Concentration of $Cl_2$ = (0.860-x) M = (0.860-0.532) M = 0.328 M

Concentration of $COCl_2$ = x M = 0.532 M

3 0

3 years ago