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sertanlavr [38]
2 years ago
10

The correctly balanced equation for H2O2 → H2O + O2 is

Physics
1 answer:
Citrus2011 [14]2 years ago
8 0

Answer:2H2O2 → 2H2O + O2

Explanation:

For a chemical equation to be balanced, the number of moles of each element at the reactant must be equal to that of the product.

According to the equation given

H2O2 → H2O + O2

There are two hydrogen atoms with two oxygen atoms at the reactant and 2hydrogen and 3oxygen at the product which shows the equation is not balanced. To balance this,

We will add 2 moles to H2O2 at the reactant and 2 moles to H2O at the product to give;

2H2O2 → 2H2O + O2 (Balanced equation)

As we can see that we now have 4 moles of Hydrogen and 4 moles of oxygen at both reactant and product making the equation balanced.

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Vince weighs 160 pounds, and his friend Nadir weighs 140 pounds. Nadir calculated that his weight on another planet would be abo
prisoha [69]

Answer:

C) 64lb

Explanation:

use the linearity method to find the weight of nadir on another planet, it is applied as follows.

Nadir Weight in earth ⇒ Nadir weight in another planet

Vince Weigh in eart  ⇒  X

our goal is to find the weight of vince in another planet (X), for this we multiply the diagonal that continents the data and divide among the remaining

140pounds    ⇒   56lb

160pounds    ⇒ X

X=\frac{(160)(56)}{140} =64lb

Vince weigh on the other planet is C) 64lb

5 0
3 years ago
Read 2 more answers
An object of mass 6 kg. is resting on a horizontal surface. A horizontal force
son4ous [18]

Answer:

a) The work done by the applied force is 1500 joules.

b) The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) 300 joules of energy are lost during motion.

Explanation:

a) Since the object has a constant mass, on which a constant horizontal force is exerted. The work done by the force (W), measured in joules, is defined by the following expression:

W = F\cdot \Delta x (1)

Where:

F - Force, measured in newtons.

\Delta x - Distance, measured in meters.

If we know that F = 15\,N and \Delta x = 100\,m, then the work done by the force exerted on the object is:

W = (15\,N)\cdot (100\,m)

W = 1500\,J

The work done by the applied force is 1500 joules.

b) At first we need to calculate the net acceleration of the object (a), measured in meters per square second. By assuming a constant acceleration, we use the following kinematic formula:

\Delta x = v_{o}\cdot t +\frac{1}{2}\cdot a\cdot t^{2} (2)

Where v_{o} is the initial velocity of the object, measured in meters per second.

We clear the acceleration within the equation above:

\frac{1}{2}\cdot a \cdot t^{2} =  \Delta x-v_{o}\cdot t

a = \frac{2\cdot (\Delta x - v_{o}\cdot t)}{t^{2}}

If we know that \Delta x = 100\,m, v_{o} = 0\,\frac{m}{s} and t = 10\,s, then the net acceleration experimented by the object is:

a = \frac{2\cdot \left[100\,m-\left(0\,\frac{m}{s} \right)\cdot (10\,s)\right]}{(10\,s)^{2}}

a = 2\,\frac{m}{s^{2}}

By the 2nd Newton's Law, we construct the following equation of equilibrium under the consideration of a friction force acting against the motion of the object:

\Sigma F = F - f = m\cdot a (3)

Where:

F - External force exerted on the object, measured in newtons.

f - Kinetic friction force, measured in newtons.

If we know that F = 15\,N, m = 6\,kg and a = 2\,\frac{m}{s^{2}}, the kinetic friction force is:

f = F-m\cdot a

f = 15\,N-(6\,kg)\cdot \left(2\,\frac{m}{s^{2}} \right)

f = 3\,N

The work done by friction (W'), measured in joules, is:

W' = f\cdot \Delta x (4)

W' = (3\,N) \cdot (100\,m)

W' = 300\,J

And the net work experimented by the object is:

\Delta W = 1500\,J - 300\,J

\Delta W = 1200\,J

By the Work-Energy Theorem we understand that change in translational kinetic energy (\Delta K), measured in joules, is equal to the change in net work. That is:

\Delta K = \Delta W (5)

If we know that \Delta W = 1200\,J, then the change in translational kinetic energy is:

\Delta K = 1200\,J

The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) The energy lost by the object is equal to the work done by the force of friction. Therefore, 300 joules of energy are lost during motion.

7 0
3 years ago
Which of the following types of light microscopy improves the resolution of thick specimens by illuminating one plane of the spe
Vilka [71]

Answer:

confocal microscopy

Explanation:

According to my research on different types of microscopes, I can say that based on the information provided within the question the tool being mentioned in this situation is a confocal microscopy. This is an extremely powerful microscope used to develop extremely sharp images of cells and tissues by viewing one plane of the specimen at a given time.

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

7 0
3 years ago
What is denser medium?​
abruzzese [7]

Answer: A medium in which speed of light is more is known as optically rarer medium and a medium in which speed of light is less is said to be optically denser medium. For example in air and water, air is raer and water is a denser medium.

Explanation:

6 0
2 years ago
A(n) 1946 kg car travels at a speed of 10 m/s . What is its kinetic energy ? Answer in units of J.
erica [24]

Answer:

KE=97300J

Explanation:

KE=1/2mv^2

KE=1/2(1946)(10)^2

KE=97300J

4 0
3 years ago
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