All categories

3 years ago

What is released during Cellular Respiration?

Chemistry

2 answers:

ValentinkaMS [17]3 years ago

8 0

Answer:

energy from glucose

Explanation:

vfiekz [6]3 years ago

6 0

Energy from glucose! brainlest please:)

You might be interested in

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy . If the rate

vovangra [49]

Answer:

K2 = 61.2 M^-1.S^-1

Explanation:

We complete the question fully:

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0kJ/mol . If the rate constant of this reaction is 6.7M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

Answer is as follows:

The question asks us to calculate the value of the rate constant at a certain temperature, given that it is at a particular value for a particular temperature. We solve the question as follows:

According to Arrhenius equation, the relationship between temperature and activation energy is as follows:

k = Ae^-(Ea/RT)

where, k = rate constant

A = pre-exponential factor

Ea = activation energy

R = gas constant

T = temperature in kelvin

From the equation, the following was derived for a double temperature problem:

ln(k2/k1) = (-Ea/R) * (1/T1 - 1/T2)

We list out the parameters as follows:

T1= (244 + 273.15) K = 517.15 K

T2= (324+ 273.15) K =597.15 K

K1 = 6.7 , K2 = ?

R = 8.314 J/mol K

Ea = 71.0 kJ/mol = 71000 J/mol

Putting the given values into the above formula as follows:

ln(k2/6.7) = (-71000/8.314) * (1/517.15 - 1/597.15)

lnk2 - 1.902 = 8539.8 * 0.000259

lnK2 = 1.902 + 2.21

lnK2 = 4.114

K2 = e^(4.114)

K2 = 61.2

Hence, K2 = 61.2 (M.S)^-1

7 0

4 years ago

Read 2 more answers

Describe the shape of the graph during the period when water is freezing. Why the graph is shaped this way?

klio [65]

Answer:

The variation of the proportion of water

Explanation:

5 0

3 years ago

Consider the above reaction using 25.0 mL of tert-butyl alcohol (d = 0.786 g/mL) with 60.0 mL of concentrated hydrobromic acid (

Liono4ka [1.6K]

Answer:

Theoretical yield: 36,3g

Percent yield: 71,9%

Explanation:

The reaction of tert-butyl alcohol with hydrobromic acid to produce tert-butyl bromide is a 1:1 reaction.

Moles of tert-butyl alcohol in 25,0 mL are:

25,0mL × (0,786g / mL) × (1mol / 74,12g) = 0,265 moles of tert-butyl alcohol

Moles of hydrobromic acid are:

60,0mL× 47,0% × (1,49g / mL) × (1mol / 80,91g) = 0,519 moles of HBr

As moles of tert-butyl alcohol are less than moles of HBr, limiting reactant is tert-butyl alcohol.

Theoretical yield of tert-butyl bromide in grams is:

0,265 moles × (137,02g / 1mol) = 36,3g

Percent yield:

26,1g / 36,3g × 100 = 71,9%

The reactions are in the image.

There are many ways to produce alkyl halides from primary and secondary alcohols. The reaction in this case is the Appel reaction.

The reaction of tertiary alkyl halides may undergo a hydrolysis reaction -under proper conditions- to form an alcohol and a hydrogen halide.

I hope it helps!

5 0

4 years ago

Which is the function of the skull

Mashcka [7]

supports the structures of the face and forms a cavity for the brain and protects the brain from harm

3 0

3 years ago

Read 2 more answers

Calculate the amount of heat, in kilojoules, required to turn solid ammonia at -78 degrees Celsius to gaseous ammonia at standar

zhuklara [117]

Answer:

132

Explanation:

3 0

4 years ago