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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

#SPJ4

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53.3% + 6.7% = 60%, 100% - 60% = 40%. 40% of glucose is made of carbon. Since there are only three types of atoms in glucose, and the amount of hydrogen and oxygen is already given, this means that whatever percentage is left (40%) has to be carbon.
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Neon is a gas made up of only one type of atom. Which term best describes neon
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The breaking buffer that we use this week contains 10mM Tris, pH 8.0, 150mM NaCl. The elution buffer is breaking buffer that als
olasank [31]

Answer:

A breakdown of the breaking buffer was first listed with its respective component and their corresponding value; then a table was made for the stock concentrations in which the volume that is being added was determined by using the formula M_1*V_1 = M_2*V_2. It was the addition of these volumes altogether that make up the 0.25 L (i.e 250 mL)  with water

Explanation:

Given data includes:

Tris= 10mM

pH = 8.0

NaCl = 150 mM

Imidazole = 300 mM

In order to make 0.25 L solution buffer ; i.e (250 mL); we have the following component.

Stock Concentration             Volume to be             Final Concentration

                                               added            

1 M Tris                                     2.5 mL                         10 mM

5 M NaCl                                  7.5 mL                        150 mM

1 M Imidazole                           75 mL                         300 mM    

M_1*V_1 = M_2*V_2. is the formula that is used to determine the corresponding volume that is added for each stock concentration

The stock concentration of Tris ( 1 M ) is as follows:

M_1*V_1 = M_2*V_2.

1*V_1= 0.01 M *250mL\\V_1 = 2.5mL

The stock concentration of NaCl (5 M ) is as follows:

M_1*V_1 = M_2*V_2.

1*V_1= 0.15 M *250mL\\V_1 = 7.5mL

The stock concentration of Imidazole (1 M ) is as follows:

M_1*V_1 = M_2*V_2.

1*V_1= 0.03 M *250mL\\V_1 = 75mL

Hence, it is the addition of all the volumes altogether that make up 0.25L (i.e 250 mL) with water.

5 0
3 years ago
How many moles are there in 3.3 grams of NH4SO2​
slavikrds [6]
1.69x10 22 molecules
8 0
3 years ago
The pressure on a 200-milliliter sample of CO2(g) at
motikmotik

Answer:

A) 100 mL

Explanation:

At constant temperature and number of moles, Using Boyle's law  

{P_1}\times {V_1}={P_2}\times {V_2}

Given ,  

V₁ = 200 mL

V₂ = ?

P₁ = 60 kPa

P₂ = 120 kPa

Using above equation as:

{P_1}\times {V_1}={P_2}\times {V_2}

{60\ kPa}\times {200\ mL}={120\ kPa}\times {V_2}

{V_2}=\frac{{60}\times {200}}{120}\ mL

{V_2}=100\ mL

3 0
3 years ago
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