2 AICI3 + 3 Ca - 3 CaCl2 + 2 Al

What is the kinetic energy of a 328 kg object that is moving at s speed of 10 m/s

Advocard [28]

16400 Joules is the kinetic energy of a 328kg object with a velocity of 10m/s.

<h3>What is the kinetic energy of the object?</h3>

Kinetic energy is simply energy possessed by a body in motion.

Kinetic energy is expressed as;

Kinetic energy = 1/2 × m × v²

Where v is velocity and m is mass of the object,

Given the data in the question;

Mass of the object m = 328kg
Velocity of the object = 10m/s
Kinetic energy = ?

To determine the kinetic energy of the object, plug the given values into the formula above.

Kinetic energy = 1/2 × m × v²

Kinetic energy = 1/2 × 328kg × (10m/s)²

Kinetic energy = 1/2 × 328kg × 100m²/s²

Kinetic energy = 328kg × 50m²/s²

Kinetic energy = 16400kgm²/s²

Kinetic energy = 16400J

Therefore, the kinetic energy 16400 Joules.

Learn more about kinetic energy here: brainly.com/question/27397088

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6 0

1 year ago

The following diagram best represents

Ilya [14]

Answer:

Mixture of two elements

Explanation:

You can tell that it is an element since the atoms are only bonded to atoms of the same type. It is a mixture since the molecules are together in solution but aren't bonded together.

4 0

3 years ago

Predict the following chemical formula for a compound between Al and S.

Alinara [238K]

Al2S3. Al has a +3 charge and S has a -2 charge, generally speaking

7 0

3 years ago

Read 2 more answers

The energy in eV for light with a wavelength of 6250 angstroms is _. Note - there are 1.6 x 10-12 erg in 1 eV.

vivado [14]

Answer:

2 eV

Explanation:

The energy of a photon of light is given by the formula

$E=\frac{hc}{\lambda}$

where

h is the Planck constant

c is the speed of light

$\lambda$ is the wavelength of the photon

In this problem we have:

$h=6.63\cdot 10^{-34} Js$

$c=3.0\cdot 10^8 m/s$

$\lambda=6250 A = 6250\cdot 10^{-10} m$ is the wavelength of the photon

Therefore, the energy in Joules is

$E=\frac{(6.63\cdot 10^{-34})(3.0\cdot 10^8)}{6250\cdot 10^{-10}}=3.2\cdot 10^{-19}J$

We want to convert this energy into electronvolts: we know that the conversion factor is

$1 eV = 1.6\cdot 10^{-19}J$

Therefore,

$E=\frac{3.2\cdot 10^{-19}}{1.6\cdot 10^{-19}}=2 eV$

5 0

3 years ago

odium carbonate (Na2CO3Na2CO3) is used to neutralize the sulfuric acid spill. How many kilograms of sodium carbonate must be add

Arte-miy333 [17]

Answer : The mass of sodium carbonate added to neutralize must be, $6.54\times 10^3kg$

Explanation :

First we have to calculate the moles of $H_2SO_4$ .

$\text{Moles of }H_2SO_4=\frac{\text{Mass of }H_2SO_4}{\text{Molar mass of }H_2SO_4}$

Given:

Molar mass of $H_2SO_4$ = 98 g/mole

Mass of $H_2SO_4$ = $6.05\times 10^3kg=6.05\times 10^6g$

Conversion used : (1 kg = 1000 g)

Now put all the given values in the above expression, we get:

$\text{Moles of }H_2SO_4=\frac{6.05\times 10^6g}{98g/mol}=6.17\times 10^4mol$

The moles of $H_2SO_4$ is, $6.17\times 10^4mol$

Now we have to calculate the moles of $Na_2CO_3$

The balanced neutralization reaction is:

$Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+H_2CO_3$

From the balanced chemical reaction we conclude that,

As, 1 mole of $H_2SO_4$ neutralizes 1 mole of $Na_2CO_3$

So, $6.17\times 10^4mol$ of $H_2SO_4$ neutralizes

Now we have to calculate the mass of $Na_2CO_3$

$\text{ Mass of }Na_2CO_3=\text{ Moles of }Na_2CO_3\times \text{ Molar mass of }Na_2CO_3$

Molar mass of $Na_2CO_3$ = 106 g/mole

$\text{ Mass of }Na_2CO_3=(6.17\times 10^4mol)\times (106g/mole)=6.54\times 10^6g=6.54\times 10^3kg$

Thus, the mass of sodium carbonate added to neutralize must be, $6.54\times 10^3kg$

3 0

4 years ago