All categories

3 years ago

What would happen if we couldn't carry respiration out?

Chemistry

1 answer:

pantera1 [17]3 years ago

8 0

Answer:

There would be no energy for cellular activities

Explanation:

Cellular respiration is the process whereby living organisms breakdown ingested food in their cells in order to obtain energy in form of ATP. In other words, respiration converts the stores energy in food molecules into usable energy by the cell. The general equation of the process is as follows:

C6H12O6 + 6O2 → 6CO2 + 6H2O

The process of cellular respiration is carried out by every living organism in order to obtain energy to perform their various metabolic activities. Hence, without RESPIRATION, there would be no energy (ATP) for use by the cell, which utterly means no cellular activities will take place.

You might be interested in

Would it be C???...idk HELP

givi [52]

Answer:

C. Butanal , is the aldehyde

Explanation:

A . It is carboxylic acid : ---COOH group

B. It is Ester : ----COOR group , Here R = CH3

C. It is Aldehyde : -----CHO group

D. It is ketone : ----C=O group

See image :

8 0

3 years ago

Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M.Express your answer to two significant f

N76 [4]

The question is incomplete, here is the complete question:

Calculate the initial rate for the formation of C at 25°C, if [A]=0.50 M and [B]=0.075 M. Express your answer to two significant figures and include the appropriate units.Consider the reaction

A + 2B ⇔ C

whose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:

The table is attached below as an image.

Answer: The initial rate for the formation of C at 25°C is $2.25\times 10^{-2}Ms^{-1}$

Explanation:

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$A+2B\rightleftharpoons C$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b$

where,

a = order with respect to A

b = order with respect to B

Expression for rate law for first trial:

$5.4\times 10^{-3}=k(0.30)^a(0.050)^b$ ....(1)

Expression for rate law for second trial:

$1.1\times 10^{-2}=k(0.30)^a(0.100)^b$ ....(2)

Expression for rate law for third trial:

$2.2\times 10^{-2}=k(0.50)^a(0.050)^b$ ....(3)

Dividing 2 by 1, we get:

$\frac{1.1\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.30)^a(1.00)^b}{(0.30)^a(0.050)^b}\\\\2=2^b\\b=1$

Dividing 3 by 1, we get:

$\frac{2.2\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.50)^a(0.050)^b}{(0.30)^a(0.050)^b}\\\\4.07=2^a\\a=2$

Thus, the rate law becomes:

$\text{Rate}=k[A]^2[B]^1$ ......(4)

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

$5.4\times 10^{-3}=k[0.30]^2[0.050]^1\\\\k=1.2M^{-2}s^{-1}$

Calculating the initial rate of formation of C by using equation 4, we get:

$k=1.2M^{-2}s^{-1}$

[A] = 0.50 M

[B] = 0.075 M

Putting values in equation 4, we get:

$\text{Rate}=1.2\times (0.50)^2\times (0.075)^1\\\\\text{Rate}=2.25\times 10^{-2}Ms^{-1}$

Hence, the initial rate for the formation of C at 25°C is $2.25\times 10^{-2}Ms^{-1}$

8 0

3 years ago

What is the total number of liters of NH3 formed when 20 liters of N2 reacts completely

goblinko [34]

Answer:

N2 + H2 ----------》NH3

On balancing it

N2. + 3.H2------->2.NH3

( 1 mol) (3 mol) (2 mol)

1 mol of nitrogen reacts with 3 mol of hydrogen to give 2 mol of ammonia.

Likewise,

20 litres of nitrogen reacts with 60 litres of hydrogen to give 40 litres of Ammonia.

Hence, the answer is 40 Litres.

5 0

3 years ago

Assuming that the experiments performed in the absence of inhibitors were conducted by adding 5 μl of a 2 mg/ml enzyme stock sol

prohojiy [21]

Hey there!:

From the given data ;

Reaction volume = 1 mL , enzyme content = 10 ug ( 5 ug in 2 mg/mL )

Enzyme mol Wt = 45,000 , therefore [E]t is 10 ug/mL , this need to be express as "M" So:

[E]t in molar = g/L * mol/g

[E]t = 0.01 g/L * 1 / 45,000

[E]t = 2.22*10⁻⁷

Vmax = 0.758 umole/min/ per mL

= 758 mmole/L/min

=758000 mole/L/min => 758000 M

Therefore :

Kcat = Vmax/ [E]t

Kcat = 758000 / 2.2*10⁻⁷ M

Kcat = 3.41441 *10¹² / min

Kcat = 3.41441*10¹² / 60 per sec

Kcat = 5.7*10¹⁰ s⁻¹

Hence kcat of xyzase is 5.7*10¹⁰ s⁻¹

Hope that helps!

4 0

3 years ago

An obese man was discovered in his air-conditioned hotel room sitting in a chair in front of the television. The air conditioner

Leno4ka [110]

Answer:

It has been approximately 6 hours after death.

Explanation:

This is because between 2-6 hours after death, the body starts becoming stiff from top to bottom, then spreading to the limbs. Since there is only rigor in his upper body, that would mean that with normal temperature and body conditions, it would be 4 or 5 hours after death. But since he is obese and in cold temperature, there is slower progression of rigor, leading to the maximum time in the first rigor mortis phase, 6 hours.

8 0

3 years ago