All categories

3 years ago

At STP, when 3.75 liters of oxygen reacts with excess glucose, what volume of carbon dioxide does it produce?

Physics

2 answers:

I am Lyosha [343]3 years ago

5 0

Answer is 3.75L. I hope this helps :)

iragen [17]3 years ago

5 0

<u>Answer:</u> The volume of carbon dioxide produced will be 3.75 L

<u>Explanation:</u>

STP conditions:

1 mole of as gas occupies 22.4 L of volume.

For the given chemical equation:

$6O_2(g)+C_6H_{12}O_6(g)\rightarrow 6H_2O(g)+6CO_2(g)$

As, glucose is present in excess, it is considered as excess reagent. Oxygen is considered as a limiting reagent.

By Stoichiometry of the reaction:

$6\times 22.4L$ of oxygen gas produces $6\times 22.4L$ of carbon dioxide gas.

So, 3.75 L of oxygen gas will produce = $\frac{6\times 22.4L}{6\times 22.4L}\times 3.75L=3.75L$ of carbon dioxide gas.

Thus, the volume of carbon dioxide produced will be 3.75 L

You might be interested in

A bag is dropped from a hovering helicopter. After 42.4s, calculate the bag’s displacement.

Ann [662]

Answer:

8,809.024m

Explanation:

Displacement , y = (gt^2/2 )

g = 9.8 m/s^2

t = 42.4s

y = (9.8 (42.4*42.4)) / 2

y = 8,809.024m

3 0

3 years ago

Give two examples of energy transforming from type to another. Make sure to tell the

pentagon [3]

Well one is from potential energy to ascensive moving energy . And another is from electrical to circuit energy.

4 0

3 years ago

Consider a wire of a circular cross-section with a radius of R = 3.17 mm. The magnitude of the current density is modeled as J =

Sloan [31]

Answer:

The current is $I = 8.9 *10^{-5} \ A$

Explanation:

From the question we are told that

The radius is $r = 3.17 \ mm = 3.17 *10^{-3} \ m$

The current density is $J = c\cdot r^2 = 9.00*10^{6} \ A/m^4 \cdot r^2$

The distance we are considering is $r = 0.5 R = 0.001585$

Generally current density is mathematically represented as

$J = \frac{I}{A }$

Where A is the cross-sectional area represented as

$A = \pi r^2$

=> $J = \frac{I}{\pi r^2 }$

=> $I = J * (\pi r^2 )$

Now the change in current per unit length is mathematically evaluated as

$dI = 2 J * \pi r dr$

Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows

$I = 2\pi \int\limits^{0.5 R}_{0} {( 9.0*10^6A/m^4) * r^2 * r} \, dr$

$I = 2\pi * 9.0*10^{6} \int\limits^{0.001585}_{0} {r^3} \, dr$

$I = 2\pi *(9.0*10^{6}) [\frac{r^4}{4} ] | \left 0.001585} \atop 0}} \right.$

$I = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ]$

substituting values

$I = 2 * 3.142 * 9.00 *10^6 * [ \frac{0.001585^4}{4} ]$

$I = 8.9 *10^{-5} \ A$

5 0

4 years ago

If a rock is dropped with a weight of 50kg and friction due to air resistance produces a force of 98N, what is the net force in

bearhunter [10]

Answer:

The net force is 392N, pointing down.

Explanation:

The net force is the sum of all forces acting on the rock, namely the gravity ("+" acting downward) and the friction force due to air resistance ("-" acting upward):

$F_{net} = F_g - F_{air}\\F_{net} = m\cdot g - F_{air} = 50kg\cdot 9.8\frac{m}{s^2}-98N= (490-98)N=392N$

The net force is 392N and pointing down (positive)/

8 0

4 years ago

The torque exerted by a crowbar on an object increases with increased _______.

Fofino [41]

Answer:

force and leverage distance

Explanation:

the formula for torque if = force x distance

(the distance above is the leverage distance on the crow bar)

therefore if there is an increase in either the torque or the leverage distance, or both, the torque exerted by the crow bar also increases.

for example

lets assume a force of 5 n is applied on the crow bar with a leverage distance of 2 m.

the torque = 5 x 2 = 10 N.m

but if the force was increased to 7 N

torque = 7 x 2 = 12 N.m

from the illustration above, we can see that the torque increased with an increase in force. There would also be an increase in torque if the distance were to be increased.

3 0

3 years ago