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3 years ago

Is the following equation balanced? 2Al + 3NiBr2 AlBr3 + 3Ni yes no

Physics

2 answers:

Oliga [24]3 years ago

8 0

Answer:

No-the equation is not balanced.

Explanation:

The balanced equation will be;

2Al + 3NiBr2→ 2AlBr3 + 3Ni

According to the law of conservation of mass, the mass of reactants should always be the same as the mass of the products in a chemical equation.

Therefore, the number of atoms of each element in a chemical equation should always be the same on both sides of the equation, that is the side of reactants and side of products.

Balancing of chemical equations ensures that the number of atoms of each element is equal in both sides of the equation

jeka943 years ago

5 0

<h3>Answer;</h3>

No-the equation is not balanced.

<h3>Explanation;</h3>

The balanced equation will be;

2Al + 3NiBr2→ 2AlBr3 + 3Ni

According to the law of conservation of mass, the mass of reactants should always be the same as the mass of the products in a chemical equation.
Therefore, the number of atoms of each element in a chemical equation should always be the same on both sides of the equation, that is the side of reactants and side of products.
Balancing of chemical equations ensures that the number of atoms of each element is equal in both sides of the equation.

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A 0.500-kg glider, attached to the end of an ideal spring with force constant undergoes shm with an amplitude of 0.040 m. comput

Nikitich [7]

There is a missing data in the text of the problem (found on internet):
"with force constant k=450N/m"

a) the maximum speed of the glider

The total mechanical energy of the mass-spring system is constant, and it is given by the sum of the potential and kinetic energy:
 $E=U+K= \frac{1}{2}kx^2 + \frac{1}{2} mv^2$
where
k is the spring constant
x is the displacement of the glider with respect to the spring equilibrium position
m is the glider mass
v is the speed of the glider at position x

When the glider crosses the equilibrium position, x=0 and the potential energy is zero, so the mechanical energy is just kinetic energy and the speed of the glider is maximum:
 $E=K_{max} = \frac{1}{2}mv_{max}^2$
Vice-versa, when the glider is at maximum displacement (x=A, where A is the amplitude of the motion), its speed is zero (v=0), therefore the kinetic energy is zero and the mechanical energy is just potential energy:
 $E=U_{max}= \frac{1}{2}k A^2$

Since the mechanical energy must be conserved, we can write
 $\frac{1}{2}mv_{max}^2 = \frac{1}{2}kA^2$
from which we find the maximum speed
 $v_{max}= \sqrt{ \frac{kA^2}{m} }= \sqrt{ \frac{(450 N/m)(0.040 m)^2}{0.500 kg} }= 1.2 m/s$

b) the speed of the glider when it is at x= -0.015m

We can still use the conservation of energy to solve this part.
The total mechanical energy is:
 $E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J$

At x=-0.015 m, there are both potential and kinetic energy. The potential energy is
 $U= \frac{1}{2}kx^2 = \frac{1}{2}(450 N/m)(-0.015 m)^2=0.05 J$
And since
 $E=U+K$
we find the kinetic energy when the glider is at this position:
 $K=E-U=0.36 J - 0.05 J = 0.31 J$
And then we can find the corresponding velocity:
 $K= \frac{1}{2}mv^2$
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.31 J}{0.500 kg} }=1.11 m/s$

c) the magnitude of the maximum acceleration of the glider;

For a simple harmonic motion, the magnitude of the maximum acceleration is given by
$a_{max} = \omega^2 A$
where $\omega= \sqrt{ \frac{k}{m} }$ is the angular frequency, and A is the amplitude.
The angular frequency is:
$\omega = \sqrt{ \frac{450 N/m}{0.500 kg} }=30 rad/s$
and so the maximum acceleration is
$a_{max} = \omega^2 A = (30 rad/s)^2 (0.040 m) =36 m/s^2$

d) the acceleration of the glider at x= -0.015m

For a simple harmonic motion, the acceleration is given by
 $a(t)=\omega^2 x(t)$
where x(t) is the position of the mass-spring system. If we substitute x(t)=-0.015 m, we find
 $a=(30 rad/s)^2 (-0.015 m)=-13.5 m/s^2$

e) the total mechanical energy of the glider at any point in its motion. 

we have already calculated it at point b), and it is given by
 $E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J$

8 0

3 years ago

The weight of a person is 500N and his foot imprint area is 0.5m^2.Calculate the total pressure exerted by person when he stands

Ghella [55]

Answer:

Pressure on both feet will be $500N/m^2$

Explanation:

Weight of the person F = 500 N

Area of foot print $A=0.5m^2$

Area of both the foot $A=2\times 0.5=1m^2$

We have to find pressure on both the feet

Pressure is equal to ratio of force and area

So pressure $P=\frac{F}{A}$

$P=\frac{500}{1}=500N/m^2$

So the pressure on both feet will be $500N/m^2$ when person stands on both feet.

7 0

3 years ago

The vector position of an object is given by r what is the torque acting on the object about the origin when a force f = (−12.5i

attashe74 [19]

Let the vector position of the object in the (x-y) plane be
$\vec{r} = x \hat{i} + y \hat{j}$

The applied force is
$\vec{f} = -12.5 \hat{i}
$

By definition, the applied torque is
$\vec{T} = \vec{r} \times \vec{f} = (x\hat{i} + y\hat{j}) \times (-12.5y \hat{i}) = 12.5\hat{k}$

Answer: $12.5y \, \hat{k}$

7 0

3 years ago

20 points Which BEST replaces the question mark? In Saudi Arabia, __________?

dsp73

The economy is based upon the production of oil. Saudi Arabi is one of the leading oil producing countries in the world. It is one of the founding counties of Opec, an oil cartel made up of middle eastern oil producing countries which try to manipulate the worldwide price of oil.

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Freud believes that defense mechanisms are reactions people have to protect themselves from uncomfortable feelings. the article

makkiz [27]

Answer:

A: Anxious

Explanation:

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