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3 years ago

To power a flashlight for 1 second, we need ____________ of energy

Chemistry

2 answers:

Doss [256]3 years ago

5 0

It is Kenetic energy

IrinaK [193]3 years ago

5 0

Answer:

Light energy

Explanation:

In a flashlight, the electrical energy becomes light energy and thermal energy in the bulb.

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A cube has a height of 8 cm and a mass of 457 g. What is its density?

ZanzabumX [31]

Answer:

the answer is b, 0.89

Explanation:

A cube has a height of 8 cm and a mass of 457 g. What is its density?

a. 233,984 g/cm

b. 0.89g/cm3

c. 1.12 g/cm3

Density = mass/volume

the volume of the cube is 8X8X8=512cm3

the mass is 457 gm

the density is 457/512 = 0.889 gm/cm3

the answer is b, 0.89

3 0

2 years ago

Balance chemical equation Fe3O4 = Fe + CO2

m_a_m_a [10]

equation Fe3O4=Fe+CO2 is an impossible reaction

5 0

4 years ago

what is the maximum number of covalent bonds that an oxygen atom with atomic number 8 can make with hydrogen? question 4 options

Oliga [24]

Maximum number of covalent bonds that an oxygen atom can make with hydrogen is 2.

the ground state electronic configuration of oxygen is 2s² 2p⁴ that means it has 6 electrons in its valence shell and require two electrons are required to complete its octate.
Two bonds are created when an electron donor atom shares the two needed electrons with oxygen. The ability of two oxygen atoms to share valence electrons results in the creation of a double bond between the two atoms.
There are no longer any empty orbitals in the octet of oxygen after it is complete. As a result, it is unable to accept more electrons or create more bonds.

Therefore, Oxygen can only generate two bonds because it needs two additional electrons to complete its octet, after which it will run out of empty orbitals in which to receive additional electrons and create additional bonds.

learn more about octate here:

https://brainly.in/question/24161245

#SPJ4

3 0

2 years ago

Calculate the number of molecules?

Ann [662]

87.33 X 10^15 molecules

8 0

3 years ago

The unit cell for cr2o3 has hexagonal symmetry with lattice parameters a = 0.4961 nm and c = 1.360 nm. If the density of this ma

IRINA_888 [86]

To calculate the packing factor, first calculate the area and volume of unit cell.

Area is calculated as:

$A=6R^{2}\sqrt{3}$

Here, R is radius and is related to a as follows:

$R=\frac{a}{2}$

Putting the value in expression for area,

$A=6(\frac{a}{2})^{2}\sqrt{3}=1.5a^{2}\sqrt{3}$

The value of a is 0.4961 nm

Since, $1 nm=10^{-7}cm$

Thus, $0.4961 nm=4.961\times 10^{-8} cm$

Putting the value,

$Area=1.5(4.961\times 10^{-8}cm)^{2}\sqrt{3}=6.39\times 10^{-15}cm^{2}$

Now, volume can be calculated as follows:

$V=Area\times c$

The value of c is 1.360 nm or $1.360\times 10^{-7} cm$

Putting the value,

$V=(6.39\times 10^{-15}cm^{2})\times (1.360\times 10^{-7} cm)=8.7\times 10^{-22}cm^{3}$

now, number of atom in unit cell can be calculated by using the following formula:

$n=\frac{\rho N_{A}V_{c}}{A}$

Here, A is atomic mass of $Cr_{2}O_{3}$ is 151.99 g/mol.

Putting all the values,

$n=\frac{(5.22 g/cm^{3})(6.023\times 10^{23} mol^{-1})(8.7\times 10^{-22}cm^{3})}{(151.99 g/mol)}\approx 18$

Thus, there will be 18 $Cr_{2}O_{3}$ units in 1 unit cell.

Since, there are 2 Cr atoms and 3 oxygen atoms thus, units of chromium and oxygen will be 2×18=36 and 3×18=54 respectively.

The atomic radii of $Cr^{3+}$ and $O^{2-}$ is 62 pm and 140 pm respectively.

Converting them into cm:

$1 pm=10^{-10}cm$

Thus,

$r_{Cr^{3+}}=6.2\times 10^{-9}cm$

and,

$r_{O^{2-}}=1.4\times 10^{-8}cm$

Volume of sphere will be sum of volume of total number of cations and anions thus,

$V_{S}=V_{Cr^{3+}}+V_{O^{2-}}$

Since, volume of sphere is $V=\frac{4}{3}\pi r^{3}$ ,

$V_{S}=36\left ( \frac{4}{3}\pi (r_{Cr^{3+})^{3}} \right )+54\left ( \frac{4}{3}\pi (r_{O^{2-})^{3}} \right )$

Putting the values,

$V_{S}=36\left ( \frac{4}{3}(3.14) (6.2\times 10^{-9} cm)^{3}} \right )+54\left ( \frac{4}{3}(3.14) (1.4\times 10^{-8} cm)^{3}} \right )=6.6\times 10^{-22}\times 10^{-8}cm^{3}$

The atomic packing factor is ratio of volume of sphere and volume of crystal, thus,

$packing factor=\frac{V_{S}}{V_{C}}=\frac{6.6\times 10^{-22}cm^{3}}{8.7\times 10^{-22}cm^{3}}=0.758$

Thus, atomic packing factor is 0.758.

6 0

3 years ago

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