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Arturiano [62]
3 years ago
8

Which of the following subatomic particles is uncharged?

Chemistry
2 answers:
juin [17]3 years ago
7 0
It’s neutrons. B. because protons are positively charged and electrons are negatively charged while neutrons are uncharged
Andrews [41]3 years ago
4 0

Answer:

d is the right one. aaßriadjcjfjfjfdjfjf

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What type of bond is between sodium and oxygen​
Greeley [361]

Answer:

Ionic

Explanation:

Sodium is Metal, Oxygen is Non-metal. Non-metals and metals are automatic Ionic bonds

3 0
3 years ago
Read 2 more answers
Compounds can be broken down by using what method?
Dima020 [189]

Answer:

chemical change

Explanation:

chemical change requires energy in the form of heat or electricity.

4 0
3 years ago
PLEASE HELP DUE TODAY 55 POINTS
Dvinal [7]

Answer : The final volume of gas will be, 26.3 mL

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 0.974 atm

P_2 = final pressure of gas = 0.993 atm

V_1 = initial volume of gas = 27.5 mL

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 22.0^oC=273+22.0=295K

T_2 = final temperature of gas = 15.0^oC=273+15.0=288K

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

\frac{0.974 atm\times 27.5 mL}{295K}=\frac{0.993 atm\times V_2}{288K}

V_2=26.3mL

Therefore, the final volume of gas will be, 26.3 mL

3 0
3 years ago
Read 2 more answers
in the reaction mgcl2 + 2koh mg(oh)2 + 2kcl, if 3 moles mgcl2 are added to 4 moles koh what determines how much mg(oh)2 is made
otez555 [7]
Every mole of MgCl2 reacts with 2 moles of KOH, therefore the 4 moles of KOH will only react with 2 moles of MgCl2, making it the limiting reagent and therefore KOH determines how much Mg(OH)2 is produced.
6 0
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:

1)

we know that

half cell with higher reduction potential is cathode

so

cathode :

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

anode :

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

so

overall reaction is

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

now

Eo cell = Eo cathode - Eo anode

so

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

so

oxidzing agents

N20 > Cr+3

reducing agents

Cr > N2

2)

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

3)

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
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