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3 years ago

What are two main forms of energy given off when paper burns and where does the energy come from

2 answers:

galina1969 [7]3 years ago

5 0

When paper burns, it releases two distinct forms of energy
Thermal Energy
Light Energy

Thermal Energy is heat energy, heat energy is formed from the flame on the paper, in this scenario.

Light energy comes from the reaction of the paper, the flame.

In conclusion, both energy's form from the reaction of the paper, and the combustion in the air. Thus, heat and light.

luda_lava [24]3 years ago

3 0

Thermal energy and light energy

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How many grams are in a sample of 0.55 mol of K?

Katyanochek1 [597]

The equation you use here is
mass =moles x Mr

So:
Moles of K - 0.55mol
Mr of K - 39.1

Mass= 0.55x39.1 =21.505g

4 0

2 years ago

5) Which of these are intensive properties?

evablogger [386]

Answer:

For number 5 it is the 2nd one some stants has a mass of 1.25 g And the 2nd 1 number 6It is the 1st 1A subject has a melting point of 40

6 0

3 years ago

Please help make sure its correct thanks

Wewaii [24]

The empirical formula of metal iodide : CoI₃(Cobalt(III) Iodide)

<h3>Further explanation</h3>

13.02 g sample of Cobalt , then mol Co(MW=58.933 g/mol) :

$\tt mol=\dfrac{mass}{Ar}=\\\\mol=\dfrac{13.02~g}{58.933}\\\\mol=0.221$

Mass of metal iodide formed : 97.12 g, so mass of Iodine :

$\tt =mass~metal~iodide-mass~Cobalt\\\\=97.12-13.02\\\\=84.1~g$

Then mol iodine (MW=126.9045 g/mol) :

$\tt \dfrac{84.1}{126.9045}=0.663$

mol ratio of Cobalt and Iodine in the compound :

$\tt 0.221\div 0.663=1\div 3$

5 0

2 years ago

What material was added to powdered rock during tuttle and bowen's experiments?

Rzqust [24]

Answer:

water was added to powdered rock

Explanation:

4 0

2 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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