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2 years ago

Compare and contrast granite and rhyolite

2 answers:

8 0

Hey there!
Great question=)

Answer: Contrast-Same chemical composition. Granite is intrusive, when rhyolite is extrusive!
Compare-Different cooling rates.

<span>I hope this helps;)</span>

Alona [7]2 years ago

5 0

Answer:

Explanation:

Hello,

In this case, we say that rhyolite is very closely related to granite, nonetheless rhyolite has a larger a mount of finer crystals than granite, which are as tiny as they are not possible to be seen by a simple look with out eyes. Moreover, rhyolite is an extrusive igneous rock that was formed by cooling down more rapidly than granite which provide it a glassy glance. Additionally, rhyolite is made up of quartz, feldspar, mica, and hornblende.

On the other hand, the granite is a shiny-like igneous rock with visible grans by a simple look (unlike rhyolite). It is formed when a sluggish crystallization of magma is carried out underground. It is majorly made up of quartz and feldspar with traces of mica, amphiboles, and other minerals.

Best regards.

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Calculate the energy (in J/atom) for vacancy formation in silver, given that the equilibrium number of vacancies at 800 C is 3.6

MAXImum [283]

Answer:

the energy vacancies for formation in silver is $\mathbf{Q_v = 3.069*10^{-4} \ J/atom}$

Explanation:

Given that:

the equilibrium number of vacancies at 800 °C

i.e T = 800°C is 3.6 x 10¹⁷ cm3

Atomic weight of sliver = 107.9 g/mol

Density of silver = 9.5 g/cm³

Let's first determine the number of atoms in silver

Let silver be represented by N

SO;

$N = \dfrac{N_A* \rho _{Ag}}{A_{Ag}}$

where ;

$N_A =$ avogadro's number = $6.023*10^{23} \ atoms/mol$

$\rho _{Ag}$ = Density of silver = 9.5 g/cm³

$A_{Ag}$ = Atomic weight of sliver = 107.9 g/mol

$N = \dfrac{(6.023*10^{23} \ atoms/mol)*( 9.5 \ g/cm^3)}{(107.9 \ g/mol)}$

N = 5.30 × 10²⁸ atoms/m³

However;

The equation for equilibrium number of vacancies can be represented by the equation:

$N_v = N \ e^{^{-\dfrac{Q_v}{KT}}$

From above; Considering the natural logarithm on both sides; we have:

$In \ N_v =In N - \dfrac{Q_v}{KT}$

Making $Q_v$ the subject of the formula; we have:

${Q_v = - {KT} In( \dfrac{ \ N_v }{ N})$

where;

K = Boltzmann constant = 8.62 × 10⁻⁵ eV/atom .K

Temperature T = 800 °C = (800+ 273) K = 1073 K

$Q _v =-( 8.62*10^{-5} \ eV/atom.K * 1073 \ K) \ In( \dfrac{3.6*10^{17}}{5.3 0*10^{28}})$

$\mathbf{Q_v = 2.38 \ eV/atom}$

Where;

1 eV = 1.602176565 × 10⁻¹⁹ J

Then

$Q_v = (2.38 \ * 1.602176565 * 10^{-19} ) J/atom }$

$\mathbf{Q_v = 3.069*10^{-4} \ J/atom}$

Thus, the energy vacancies for formation in silver is $\mathbf{Q_v = 3.069*10^{-4} \ J/atom}$

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If 72.5 grams of calcium metal (Ca) react with 65.0 grams of oxygen gas (O2) in a synthesis reaction, how many grams of the exce

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2Ca + O2 = 2CaO
First, determine which is the excess reactant
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Since the ratio of to O2 is 2:1 in the balanced reaction, divide Ca's molar mass by 2 to get 0.9044862518. this isn't necessary because Ca is already obviously the limiting reactant. therefore, O2 is the excess reactant.

Now do the stoichiometry
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