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4 years ago

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Which of the following is a balanced chemical equation? MgF 2 + Li 2CO 3 ⟶ MgCO 3 + LiF 2Al + 6HCl ⟶ 3H 2 + AlCl 3 P 4 + 3O 2 ⟶

2P 2O 3 CH 4 + 2O 2 ⟶ CO 2 + H

1 answer:

IRISSAK [1]4 years ago

3 0

P₄ + 3O₂ ⟶ 2P₂O₃

Explanation:

A balanced equation is any equation that complies with the law of conservation of matter and mass.

The law of conservation of matter states that "in a chemical reaction, matter is neither created nor destroyed but simply rearranges. ".

This suggests that the total number of atoms on both reactant and product sides must be the same in a balanced chemical equation. Also, the same atoms are found on both sides of a chemical equation with no additional or external species.

Now let us check if the equation is conserved:

P₄ + 3O₂ ⟶ 2P₂O₃

Atoms reactant product

P 4 4

O 6 6

We see that we have equal number of atoms on both sides. The equation is balanced.

learn more:

Balanced equation brainly.com/question/2612756

#learnwithBrainly

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At which life cycle stage are stars most stable

asambeis [7]

~Hello there!

Your question: At which life cycle stage are stars most stable?

Your answer: The young star stage is the life cycle where the stars are most stable because the stars are just flaming, no interference nor there is no disturbance.

Any queries ^?

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4 0

3 years ago

Read 2 more answers

A uniform rod of mass 3.30×10−2 kg and length 0.450 m rotates in a horizontal plane about a fixed axis through its center and pe

Alex73 [517]

(a) 2.75 rev/min

The moment of inertia of the rod rotating about its center is:

$I_R=\frac{1}{12}ML^2$

where

$M=3.30\cdot 10^{-2} kg$ is its mass

L = 0.450 m is its length

Substituting,

$I_R=\frac{1}{12}(3.30\cdot 10^{-2})(0.450)^2=5.57\cdot 10^{-4} kg m^2$

The moment of inertia of the two rings at the beginning is

$I_r = 2mr^2$

where

m = 0.200 kg is the mass of each ring

$r=5.20\cdot 10^{-2} m$ is their distance from the center of the rod

Substituting,

$I_r=2(0.200)(5.20\cdot 10^{-2})^2=1.08\cdot 10^{-3} kg m^2$

So the total moment of inertia at the beginning is

$I_1=I_R+I_r = 5.57\cdot 10^{-4}+1.08\cdot 10^{-3}=1.64\cdot 10^{-3}kg m^2$

The initial angular velocity of the system is

$\omega_1 = 35.0 rev/min$

The angular momentum must be conserved, so we can write:

$L=I_1 \omega_1 = I_2 \omega_2$ (1)

where $I_2$ is the moment of inertia when the rings reach the end of the rod; in this case, the distance of the ring from the center is

$r=\frac{0.450 m}{2}=0.225 m$

so the moment of inertia of the rings is

$I_r=2(0.200)(0.225)^2=0.0203 kg m^2$

and the total moment of inertia is

$I_2 = I_R + I_r =5.57\cdot 10^{-4} + 0.0203 = 0.0209 kg m^2$

Substituting into (1), we find the final angular speed:

$\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{0.0209}=2.75 rev/min$

(b) 103.0 rev/min

When the rings leave the rod, the total moment of inertia is just equal to the moment of inertia of the rod, so:

$I_2 = I_R = 5.57\cdot 10^{-4}kg m^2$

So using again equation of conservation of the angular momentum:

$L=I_1 \omega_1 = I_2 \omega_2$

We find the new final angular speed:

$\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{5.57\cdot 10^{-4}}=103.0 rev/min$

7 0

3 years ago

Four pairs of objects have the masses as described below, along with the distances between

lord [1]

Answer:

<h2>Mass of 1 Kg and 2 Kg, 1 meters apart.</h2>

Explanation:

The gravitational force is defined as

$F=G\frac{m_{1} m_{2} }{r^{2} }$

By definition, the gravitational force depends directly on the product of the masses and indirectly on the distance between the masses, which means the further they are, the less gravitational force would be. And, the greater the masses, the greater the gravitational force.

Among the options, the pair that would have the greatest gravitational force is Mass of 1 Kg and 2 Kg, with 1 meter between them.

Notice that the last choice includes the same masses but with a greater distance between them, that means it would be a weaker graviational force.

Therefore, the right answer is the second choice.

7 0

3 years ago

The diameter of a copper atom is approximately 2.28e-10 m. The massof one mole of copper is 64 grams. Assume that the atoms area

KIM [24]

1) Mass of one copper atom: $1.063\cdot 10^{-22} kg$

2) There are $9.33\cdot 10^{24}$ atoms in the cube

3) Mass of the cubical block: 992 kg

Explanation:

1)

We are told here that the mass of one mole of copper is

$M=64 g$

for

$n=1 mol$ (number of moles)

We also know that the number of atoms inside 1 mole of substance is equal to Avogadro number:

$N_A = 6.022\cdot 10^{23}$

This means that $N_A$ atoms of copper have a mass of M = 64 g. Therefore, we can find the mass of one copper atom by dividing the total mass by the avogadro number:

$m=\frac{M}{N_A}=\frac{64}{6.022\cdot 10^{23}}=1.063\cdot 10^{-22} kg$

2)

We are told that the diameter of a copper atom is

$d=2.28\cdot 10^{-10} m$

We can assume that the atoms are arranged in a cube, and that they are all attached to each other; so the side of the cube can be written as size of one atom multiplied by the number of atom per side:

$L=Nd$

where

N is the number of atoms (rows) in one side of the cube

Since the side of the cube is

L = 4.8 cm = 0.048 m

We find N:

$N=\frac{L}{d}=\frac{0.048}{2.28\cdot 10^{-10}}=2.11\cdot 10^8$

This is the number of atom rows per side; therefore, the total number of atoms in the cube is

$N^3=(2.11\cdot 10^8)^3=9.33\cdot 10^{24}$

3)

The total mass of the cubical block of copper will be given by the mass of one atom of copper multiplied by the total number of atoms, so:

$M= N^3 m$

where:

$N^3 = 9.33\cdot 10^{24}$ is the number of atoms in the cube

$m=1.063\cdot 10^{-22} kg$ is the mass of one atom

Therefore, substituting, we find:

$M=(9.33\cdot 10^{24})(1.063\cdot 10^{-22})=992 kg$

So, the mass of the cubical block is 992 kg.

Learn more about mass and density:

brainly.com/question/5055270

brainly.com/question/8441651

#LearnwithBrainly

3 0

3 years ago

A subway train accelerates from rest at one station at a rate of 1.30 m/s^2 for half of the distance to the next station, then d

Minchanka [31]

This problems a perfect application for this acceleration formula:

   Distance = (1/2) (acceleration) (time)² .

During the speeding-up half:   1,600 meters = (1/2) (1.3 m/s²) T²
During the slowing-down half: 1,600 meters = (1/2) (1.3 m/s²) T²

Pick either half, and divide each side by 0.65 m/s²:

   T² = (1600 m) / (0.65 m/s²)

   T = square root of (1600 / 0.65) seconds

Time for the total trip between the stations is double that time.

   T = 2 √(1600/0.65) = <em>99.2 seconds</em> (rounded)

7 0

3 years ago