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Alex
2 years ago
13

if a substance causes another substance to be oxidized, that substance is a(n) agent. likewise, the substance that is oxidized i

n an electrochemical reaction is always the agent.
Chemistry
1 answer:
Inessa [10]2 years ago
6 0

If a substance causes another substance to be oxidized, that substance is an oxidizing agent. Likewise, the substance that is oxidized in an electrochemical reaction is always the reducing agent.

An Oxidizing agent is the one that gets reduced by accepting electrons. It causes oxidation because it makes the other substance lose electrons.

Oxidizing agents also transfer one electronegative atom like oxygen to the other chemical substance. Halogens are an example of oxidizing agents.

A reducing agent is the one that gets oxidized because it loses electrons in a redox reaction. It loses electrons and achieve a higher oxidation state. Lithium is an example of a reducing agent.

If you need to learn more about oxidizing and reducing agent, click here

brainly.com/question/20565173?referrer=searchResults

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Which event is an example of melting?
avanturin [10]

Answer:

A. Wax drips down the side of a lot candle.

Explanation:

The chemical change from solid to liquid. This is a combustion reaction, so carbon dioxide gas and water vapour is also produced but you can't see them

8 0
3 years ago
18. When a chemical reaction takes place in an open system, matter_____.
zheka24 [161]
A) can enter from the surroundings, but cannot escape to the surroundings
4 0
3 years ago
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What is the temperature of 0.500 moles of helium that occupies a volume of 15.0 L at a pressure of 1.6 atm?
denis23 [38]

Answer:

600K

Explanation:

PV=nRT

T=PV/nR

= 1.6atm* 15.0L/ 0.5mol*0.0821LatmK^-1mol^-1

=600K

6 0
3 years ago
A sample with a molar mass of 34.00 g/mol is found to consist of 5.9783% Hydrogen and 94.0217% oxygen. Find it’s molecular formu
kozerog [31]

Answer:

The formula is H202 (hydrogen peroxide, known as hydrogen peroxide)

Explanation:

100%----34g

5, 9783%---X= (5, 9783%x 34g)/100% =2 g

1g---1 atom of H

2g----x= 2g x 1 atom of H/1g = 2 atom of H

100%----34g

94, 0217%---X= (94,0217%x 34g)/100% =32 g

16g---1 atom of 0

32g----x= 32g x 1 atom of 0/16g = 2 atom of 0

5 0
2 years ago
Using the standard enthalpies of formation found in the textbook, determine the enthalpy change for the combustion of ethanol c2
ArbitrLikvidat [17]
Enthalpy of formation is calculated by subtracting the total enthalpy of formation of the reactants from those of the products. This is called the HESS' LAW.
ΔHrxn = ΔH(products) - ΔH(reactants)

Since the enthalpies are not listed in this item, from reliable sources, the obtained enthalpies of formation are written below.
ΔH(C2H5OH) = -276 kJ/mol
ΔH(O2) = 0 (because O2 is a pure substance)
ΔH(CO2) = -393.5 kJ/mol
ΔH(H2O) = -285.5 kJ/mol

Using the equation above,
ΔHrxn = (2)(-393.5 kJ/mol) + (3)(-285.5 kJ/mol) - (-276 kJ/mol)
ΔHrxn = -1367.5 kJ/mol

<em>Answer: -1367.5 kJ/mol</em>
6 0
3 years ago
Read 2 more answers
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