The cellular process that breaks down glucose to produce energy is_____

Which substance loses electrons in a chemical reaction? the one that is oxidized, which is the oxidizing agent the one that is r

victus00 [196]

Answer:

The one that is oxidized or the reducing agent

Explanation:

Oxidation results in loss of electrons or an increase in oxidation state.

Reduction results in gain of electrons or decrease in oxidation state.

The element that undergoes oxidation is said to oxidised, similarly the element that undergoes reduction is said to be reduced.

In a redox reaction, both oxidation and reduction takes place.

If a substrate undergoes oxidation or is oxidised, it is also responsible for reduction of the other species, as the total number of electrons should always be conserved.

The substance that undergoes oxidation, releases some electrons, these electrons are taken by the other substrate and it undergoes reduction.

Hence the substance that undergoes oxidation is called the reducing agent, as it is responsible for reduction of other substrate, when oxidizing itself.

Similar thought works for a substance that undergoes reduction or is reduced, works as an oxidizing agent.

7 0

3 years ago

Read 2 more answers

Plz help.chemissttrryy<br><br><br><br>

Gelneren [198K]

Attached is the answer

3 0

3 years ago

The raw water supply for a community contains 18 mg/L total particulate matter. It is to be treated by addition of 60 mg alum (A

s344n2d4d5 [400]

Solution :

Given :

The steady state flow = 8000 $m^3 /d$

$= 80 \times 10^5 \ I/d$

The concentration of the particulate matter = 18 mg/L

Therefore, the total quantity of a particulate matter in fluid $= 80 \times 10^5 \ I/d \times 18 \ mg/L$

$= 144 \times 10^6 \ mg/g$

$= 144 \ kg/d$

If 60 mg of alum $[Al_2(SO_4)_3.14 H_2O]$ required for one litre of the water treatment.

So Alum required for $80 \times 10^5 \ I/d$

$= 80 \times 15^5 \ I/d \times 60 \ mg \ alum /L$

$= 480 \times 10^6 \ mg/d$

or 480 kg/d

Therefore the alum required is 480 kg/d

1 mg of the alum gives 0.234 mg alum precipitation, so 60 mg of alum will give $= 60 \times 0.234 \text{ of alum ppt. per litre}$

$= 14.04$ mg of alum ppt. per litre

480 kg of alum will give = 480 x 0.234 kg/d

= 112.32 kg/d ppt of alum

Daily total solid load is $= 144 \ kg/d + 112.32 \ kg/d$

= 256.32 kg/d

So, the total concentration of the suspended solid after alum addition $= 18 \ mg/L + 60 \times 0.234$

= 32.04 mg/L

Therefore total alum requirement = 480 kg/d

b). Initial pH = 7.4

The dissociation reaction of aluminium hydroxide as follows :

$Al(OH)_3 \rightleftharpoons Al^{3+} + 3OH^{-}$

After addition, the aluminium hydroxide pH of water will increase due to increase in $OH^-$ ions.

Therefore, the pH of water will be acceptable range after the addition of aluminium hydroxide.

c). The reaction of $CO_2$ and water as follows :

$CO_2 (g) + H_2O (l) \rightarrow H_2CO_3$

For the atmospheric pressure :

$p_{CO_2} = 3.5 \times 10^{-4} \ atm$

And the pH is reduced into the range of 5.9 to 6.4

6 0

3 years ago

ubstance A undergoes a first order reaction A®B with a half-life, t½, of 20 min at 25 °C. If the initial concentration of A in a

Stells [14]

Answer : The concentration of A after 80 min is, 0.100 M

Explanation :

Half-life = 20 min

First we have to calculate the rate constant, we use the formula :

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{20\text{ min}}$

$k=3.465\times 10^{-2}\text{ min}^{-1}$

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $3.465\times 10^{-2}\text{ min}^{-1}$

t = time passed by the sample = 80 min

a = initial amount of the reactant = 1.6 M

a - x = amount left after decay process = ?

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

$80=\frac{2.303}{3.465\times 10^{-2}}\log\frac{1.6}{a-x}$

$a-x=0.100M$

Therefore, the concentration of A after 80 min is, 0.100 M

3 0

3 years ago

What is defined as the basic unit in the metric system that describes a quality?

Taya2010 [7]

Answer:

Explanation:

The metric system is a system of measurement that uses the meter, liter, and gram as base units of length (distance), capacity (volume), and weight (mass) respectively.

To measure smaller or larger quantities, we use units derived from the metric units

metric-system

The given figure shows the arrangement of the metric units, which are smaller or bigger than the base unit.

The units to the right of the base unit are smaller than the base unit. As we move to the right, each unit is 10 times smaller or one-tenth of the unit to its left. So, a ‘deci’ means one-tenth of the base unit, ‘centi’ is one-tenth of ‘deci’ or one-hundredth of the base unit and ‘milli’ is one-tenth of ‘centi’ or one-thousandth of the base unit.

The units to the left of the base unit are bigger than the base unit. As we move to the left, each unit is 10 times greater than the unit to its right. So, a ‘deca’ means ten times of the base unit, ‘hecto’ is ten times of ‘deca’ or hundred times of the base unit and ‘killo’ is ten times of ‘hecto’ or thousand times of the base unit.

8 0

3 years ago

The cellular process that breaks down glucose to produce energy is______.