All categories

3 years ago

If you can answer as many of these as possible that would be great, will give thanks, 5 stars, and brainliest :)

Chemistry

1 answer:

alexandr1967 [171]3 years ago

3 0

Answer:

The nucleus of the atom as suggested by "nuclear" chemistry.
Radioactive element are generally very large due to both their high number of protons, neutrons, and electrons.
An excessive amount of protons or neutrons.
Periodic table, table O, and table N.
Personal opinion question.
The atomic number.
It represents the number of protons, and can also represent the number of electrons in a neutral atom (an atom with no charge).
Protons ---- element.
Random, self-generated.
Gamma radiation is the only thing on table O that is mass less. Close to being the same is Beta particles which have almost no mass.
Neutrons and Gamma radiation.
Gamma radiation.
Electrons and Beta particles.
To turn one element into the isotope of another. Yes, I believe it is aptly named because you are changing the identity of the element.
The number of protons because they are what define the identity of an element.

Goodluck!

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Which of the following is not equal 225 g?

vekshin1

G to dkg is 1 place to the right, move the decimal point 1 place to the left
225 g = 22.5 dkg
C is the answer

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4 years ago

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What is the scientific study of how matter and energy interact

spayn [35]

Answer: Hope This Helps!

Explanation:

Physics is the scientific study of matter and energy and how they interact with each other. This energy can take the form of motion, light, electricity, radiation, gravity — just about anything, honestly.

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3 years ago

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Use the following half-reactions to write three spontaneous reactions, calculate E°cell for each reaction, and rank the oxidizin

Ainat [17]

Answer:

See explaination

Explanation:

we know that

half cell with higher reduction potential is cathode

cathode :

N20 + 2H+ + 2e- ---> N2 + H20

anode :

Cr(s) ---> Cr+3 + 3e-

overall reaction is

3 N20 + 6H+ + 2 Cr ---> 3N2 + 3H20 + 2Cr+3

now

Eo cell = Eo cathode - Eo anode

EO cell = 1.77 + 0.74

Eo cell = 2.51 V

now

in this case

oxidizing agents are N20 and Cr+3

reducing agents are Cr and N2

higher the reduction potential , stronger the oxidizing agent

lower the reduction potential , stronger the reducing agent

oxidzing agents

N20 > Cr+3

reducing agents

Cr > N2

cathode :

Au+ + e- --> Au

anode :

Cr ---> Cr+3 + 3e-

overall reaction

3Au+ + Cr ---> 3Au + Cr+3

Eo cell = 1.69 + 0.74

Eo cell = 2.43

now

oxidizing agents :

Au+ > Cr+3

reducing agents :

Cr > Au

cathode :

N20 + 2H+ + 2e- ---> N2 + H20

andoe :

Au ---> Au+ + e-

overall

2 Au + N20 + 2H+ --> 2 Au+ + N2 + H20

Eo cell = 1.77 - 1.69

Eo cell = 0.08

oxidizing agents

N20 > Au+

reducing agents

Au > N2

8 0

3 years ago

How many milliliters are in one milligram?

mariarad [96]

1 milliliters in one milligram.

8 0

3 years ago

Given the following data: Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) ΔH = −23 kJ 3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g) ΔH = −39

nalin [4]

Answer:

ΔH° = -11 kj

Explanation:

Step 1: Data given

1) Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) ΔH = −23 kJ

2) 3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g) ΔH = −39 kJ

3) Fe3O4(s) + CO(g) → 3 FeO(s) + CO2(g) ΔH = +18 kJ

Step 2: The balanced equation

FeO + CO → Fe + CO2

Step 3: Calculate ΔH for the reaction FeO(s) + CO(g) → Fe(s) + CO2(g).

To get this equation, we need to combine the 3 equations

We have to multiply the third equation by 2.

2Fe3O4(s) + 2CO(g) → 6 FeO(s) + 2CO2(g) ΔH = +54 kJ

This equation we add to the second equation

3Fe2O3(s) + CO(g) + 2Fe3O4(s) + 2CO(g) → 2Fe3O4(s) + CO2(g) + 6FeO(s) + 2CO2 (g)

3Fe2O3(s) + 3CO(g) → 3CO2(g) + 6FeO(s)

ΔH° = 2*18 + (-39) = 36 - 39 = -3 kJ

This new equation we will divide by 3

3Fe2O3(s) + 3CO(g) → 3CO2(g) + 6FeO(s)

Fe2O3(s) + CO(g) → CO2(g) + 2FeO(s)

ΔH° =-3/3 = -1 kJ

Now we will substract this new equation from the first equation

Fe2O3(s) + 3CO(g) - Fe2O3(s) - CO(g) → 2Fe(s) + 3CO2(g) -2FeO(s) - CO2(g)

2CO(g) + 2FeO(s) → 2Fe(s) + 2CO2(g)

ΔH° = -23kJ +1kJ

ΔH° = -22 kj

The next equation we will divide by 2

2CO(g) + 2FeO(s) → 2Fe(s) + 2CO2(g)

CO(g) + FeO(s) → Fe(s) + CO2(g)

ΔH° = -22kJ /2

ΔH° = -11 kj

7 0

3 years ago