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3 years ago

Quiz

Chemistry

1 answer:

Ierofanga [76]3 years ago

8 0

Answer:

It has varying degrees of thickness.

Explanation:

The oceanic crust of earth is different from the continental part of the earth's crust. It is much thinner which is about 5 km to 80 km in thickness. It is described as having much similarities with the mantle of the earth which is also very varying in thickness.

Explanation:

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Which element has similar chemical properties compared to Carbon (C)?

Dafna11 [192]

Answer:

silicon (S)

Explanation:

5 0

3 years ago

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

8 0

3 years ago

Find concentration of solution of 45g of glucose is dissolved in water to prepare 500g solution?

ivolga24 [154]

Answer:

9% solution by mass

Explanation:

If there is 500 gm of solution and 45 g of it is glucose then:

45/500 * 100% = 9 % solution

3 0

2 years ago

Calculate ΔH o rxn for the following: CH4(g) + Cl2(g) → CCl4(l) + HCl(g)[unbalanced] ΔH o f [CH4(g)] = −74.87 kJ/mol ΔH o f [CCl

anzhelika [568]

Answer: ΔH for the reaction is -277.4 kJ

Explanation:

The balanced chemical reaction is,

$CH_4(g)+Cl_2(g)\rightarrow CCl_4(l)+HCl(g)$

The expression for enthalpy change is,

$\Delta H=\sum [n\times \Delta H(products)]-\sum [n\times \Delta H(reactant)]$

$\Delta H=[(n_{CCl_4}\times \Delta H_{CCl_4})+(n_{HCl}\times B.E_{HCl}) ]-[(n_{CH_4}\times \Delta H_{CH_4})+n_{Cl_2}\times \Delta H_{Cl_2}]$

where,

n = number of moles

Now put all the given values in this expression, we get

$\Delta H=[(1\times -139)+(1\times -92.31) ]-[(1\times -74.87)+(1\times 121.0]$

$\Delta H=-277.4kJ$

Therefore, the enthalpy change for this reaction is, -277.4 kJ

4 0

3 years ago

A student puts 0.020 mol of methyl methanoate into an empty and rigid 1.0 L vessel at 450 K. The pressure is measured to be 0.74

stellarik [79]

Explanation:

Starting moles of ethanol acid = 0.020 mol

At the equilibrium 50 % of the ethanol acid molecules reacted

∴ Moles of ethanol acid reacted = 0.020 mol * 50 %/100 %

= 0.010 mol

Moles of ethanol acid remain = 0.020 mol + 0.010 mol = 0.010 mol

Moles of the product $(CH3COOH)^{2}$ gas formed are calculated as

0.010 mol CH3COOH * 1 mol $(CH3COOH)^{2}$ / 2 mol CH3COOH

= 0.005 mol $(CH3COOH)^{2}$

Therefore at the equilibrium total moles of gas present in the vessel are 0.010 mol CH3COOH and 0.005 mol $(CH3COOH)^{2}$

That is total gas moles at equilibrium = 0.010 mol + 0.005 mol = 0.015 mol

Now Calculate the pressure :

0.020 mol gas has pressure of 0.74 atm therefore at the same condition what will be the pressure exerted by 0.015 mol gas

P1/n1 = P2/n2

P2 = P1*n2 / n1

= 0.74 atm * 0.015 mol / 0.020 mol

= 0.555 atm

4 0

3 years ago