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4 years ago

12

One molecule of glucose makes 30 molecules of atp. how many molecules of glucose are needed to make 300 molecules of atp in aero

bic respiration?

2 answers:

leonid [27]4 years ago

8 0

10 molecules are required

abruzzese [7]4 years ago

4 0

Answer:

10

Explanation:

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Here is something to blow your mind. In one slice of "everything" pizza there are 650 Calories (1000 chemistry calories). That e

amid [387]

Answer:

Number of slices of pizza is 993.

Explanation:

It is given that, in one slice of "everything" pizza there are 650 Calories. The conversion factor from calories to joules is :

1 calorie = 4.184 joules

650 calories = 2719.6 joules

Total energy in the pizza, E = 2700000 J

Let there are n number of slices of pizza. It is given by :

$n=\dfrac{2700000}{2719.6}$

n = 992.79

or

n = 993

So, there are 993 slices of pizza. Hence, this is the required solution.

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3 years ago

Most binary systems with an invisible companion contain a large, bright star and a small, dim star hidden by the light of its la

Vlada [557]

Answer:

Brain signals are converted

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4 years ago

Using the conversion factor from eV (electron volts) to joules, determine which energy line for an electron dropping from one en

Tom [10]

Explanation:

Wavelength in an emission spectrum, $\lambda=435\ nm=435\times 10^{-9}\ m$

The energy of an electron is given by :

$E=\dfrac{hc}{\lambda}$

Where

h is the Planck's constant

c is the speed of light

For 435 nm, the energy of the electron will be :

$E=\dfrac{6.63\times 10^{-34}\times 3\times 10^8\ m/s}{435\times 10^{-9}}$

$E=4.57\times 10^{-19}\ J$

We know that $1\ eV=1.6\times 10^{-19}\ J$

So, $E=\dfrac{4.57\times 10^{-19}}{1.6\times 10^{-19}}$

So, E = 2.86 eV

The energy of the electron dropping from one energy level is 2.86 eV. We know that,

$\dfrac{hc}{\lambda}=E_{n_2}-E_{n_1}$

From the given energy levels :

$E_5-E_2=-0.544-(-3.403)$

$E_5-E_2=2.859\ eV$

So, the transition must be from E₅ to E₂. Hence, this is the required solution.

5 0

4 years ago

Read 2 more answers

A student is trying to determine the acceleration of a feather as she drops it to the ground. if the student is looking to achie

Anna [14]

The coordinate system should have the origin at the point where the feather is dropped and the downward direction is to be taken as positive.

All falling bodies experience acceleration towards the center of the Earth due to the force of gravitational attraction exerted on the object by the Earth. A feather, when dropped experiences an acceleration in the downward direction. Since the acceleration of the feather is in the downward direction, a feather, when dropped with zero initial velocity, has its velocity vector directed in the direction of its acceleration.

If the downward direction is taken as positive, the falling feather can be said to have a positive velocity and a positive acceleration.

5 0

3 years ago

An electron remains suspended between the surface of the Earth (assumed neutral) and a fixed positive point charge, at a distanc

DaniilM [7]

To solve this problem we will apply the concept related to the balance of forces. Here the electrostatic force defined by Coulomb's laws must be proportional to the weight of the electron, defined by Newton's second law, so that such balance exists, both can be described as:

$F_e = \frac{kq_1q_2}{d^2}$

Here,

$q_1 =$ Charge required

$q_2 =$ Charge of electron

d = Distance

k = Coulomb's constant

$F_w$ = mg

Here,

m = mass

g = Gravitational acceleration

In equilibrium the sum of Force is zero, then,

$\sum F = 0$

$F_e = F_w$

$\frac{kq_1q_2}{d^2} = mg$

$\frac{(8.98755*10^9 )(q_1)(1.6*10^{-19})}{8.8^2} = (9.8)(9.10939*10^{-31} )$

$q_1 = 4.8075*10^{-19}C$

Therefore the charge required for this to happen is $4.8075*10^{-19}C$

5 0

3 years ago