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2 years ago

which one of the following groups are decomposers a. algae b. protist c. fungi d. green plants e. photosynthetic bacteria

Chemistry

1 answer:

alukav5142 [94]2 years ago

7 0

Answer:c

Explanation:

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The electron configuration belonging to the atom with the highest second ionization energy is ________.

Gemiola [76]

Answer:

Element Lithium

Explanation:

The element with the highest second ionization energy is lithium. It belongs to the alkaline metal group I.e group one metals

It has the highest second ionization energy because it is very difficult to remove the electron from the 1s orbital.

Its atomic number is 3. The electronic configuration is 1s2 2S1

5 0

3 years ago

The 1995 Nobel Prize in Chemistry was shared by Paul Crutzen, F. Sherwood Rowland, and Mario Molina for their work concerning th

horrorfan [7]

Answer:

The enthalpy of reaction for the reaction of chlorine with ozone is -162.5 kJ.

Explanation:

$ClO ( g ) + O_3 ( g )\rightarrow Cl ( g ) + 2 O_2 ( g ),\Delta H^o_{1,rxn} =-122.8 kJ$ ..[1]

$2 O_3 ( g )\rightarrow 3O_2 ( g ),\Delta H^o_{2,rxn} = -285.3 kJ$ ..[2]

$O_3(g) + Cl(g)\rightarrow ClO (g)+O_2(g),\Delta H^o_{3,rxn}=?$ ..[3]

The enthalpy of reaction for the reaction of chlorine with ozone can be calculated by using Hess's law:

[2] - [1] = [3]

$\Delta H^o_{3,rxn}=\Delta H^o_{2,rxn}-\Delta H^o_{1,rxn}$

$=-285.3 kJ-(-122.8 kJ)=162.5 kJ$

The enthalpy of reaction for the reaction of chlorine with ozone is -162.5 kJ.

8 0

3 years ago

Good morning, I have a question..

matrenka [14]

Answer:

<h3>The answer is 5.24 mL</h3>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

$volume = \frac{mass}{density} \\$

From the question

mass = 152 g

density = 29 g/cm³

We have

$volume = \frac{152}{29} \\ = 5.24137931...$

We have the final answer as

Hope this helps you

7 0

3 years ago

What interactions are responsible for maintaining quaternary protein structure? Select all that apply.

sp2606 [1]

The ones that apply are A and C

4 0

3 years ago

Determine the number of molecules in a 100. gram sample of CCl4

enyata [817]

100. g CCl4* (1 mol CCl4/ 153.8 g CCl4)* (6.02*10^23 CCl4 molecules/ 1 mol CCl4)= 3.91*10^23 CCl4 molecules.
(Note that the units cancel out so you get the answer)

Hope this helps~

8 0

3 years ago