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3 years ago

11 points!!! what happened when glucose reacts with air?

1 answer:

3 0

When glucose (C6H12O6) reacts with oxygen, carbon dioxide and water are produced. The reaction produces moles of carbon dioxide and moles of water.

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3 years ago

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20 POINTS!!!

hram777 [196]

I think its inductance. If its not then I think its none of the above

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3 years ago

How is the mass number calculated for an element?

natali 33 [55]

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4 years ago

A horizontal force of 750 N is needed to overcome the force of static friction between a level floor and a 250-kg crate. What is

loris [4]

Answer:

The acceleration of the crate is $1.82\ m/s^2$ .

Explanation:

Given that,

Force, F = 750 N

Mass of the crate, m = 250 kg

The coefficient of friction is 0.12.

We need to find the acceleration of the crate. The net force acting on the crate is given by :

$F=ma\\\\F-f=ma$

f is frictional force, $f=\mu N=\mu mg$

$F-\mu mg=ma\\\\a=\dfrac{F-\mu mg}{m}\\\\a=\dfrac{750-0.12\times 250\times 9.8}{250}\\\\a=1.82\ m/s^2$

So, the acceleration of the crate is $1.82\ m/s^2$ . Hence, this is the required solution.

4 0

3 years ago

Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e

Nataliya [291]

Answer:

$q=1.95*10^{-7}C$

Explanation:

According to the free-body diagram of the system, we have:

$\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)$

So, we can solve for T from (1):

$T=\frac{mg}{cos(15^\circ)}(3)$

Replacing (3) in (2):

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e$

The electric force ( $F_e$ ) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)$

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

$sin(15^\circ)=\frac{\frac{r}{2}}{L}\\r=2Lsin(15^\circ)(5)$

Finally, we replace (5) in (4) and solving for q:

$mgtan(15^\circ)=\frac{kq^2}{(2Lsin(15^\circ))^2}\\q=\sqrt{\frac{mgtan(15^\circ)(2Lsin(15^\circ))^2}{k}}\\q=\sqrt{\frac{10*10^{-3}kg(9.8\frac{m}{s^2})tan(15^\circ)(2(0.22m)sin(15^\circ))^2}{8.98*10^{9}\frac{N\cdot m^2}{C^2}}}\\q=1.95*10^{-7}C$

5 0

4 years ago

11 points!!! what happened when glucose reacts with air? ​

11 points!!! what happened when glucose reacts with air?