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2 years ago

Fe(NO3)2 not sure how to get the oxidation numbers of all elements

1 answer:

3 0

<span>Fe(NO3)2
The NO3 part is a poly-atomic ion with total charge -1.
This is because Fe has a +2 charge and two NO3's with a -1 charge will balance out to 0.
Most often we just make the assumption that Oxygen has a -2 oxidation number because it is very electro-negative.
So to find N, we just need an oxidation number that balances out with 3(-2) to get -1 (the total charge of the ion)</span>

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What do nucleic acids and phospholipids have in common?

erastova [34]

They are both important cell signaling molecules and both of them contain phosphate groups.

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2 years ago

Calculate the ΔHrxn for the following

seropon [69]

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2 years ago

A certain chemical reaction releases 36.2 kJ/g of heat for each gram of reactant consumed. How can you calculate what mass of re

Lilit [14]

Answer:

0.038 g of reactant

Explanation:

Data given:

Heat release for each gram of reactant consumption = 36.2 kJ/g

mass of reactant that release 1360 J of heat = ?

Solution:

As 36.2 kJ of heat release per gram of reactant consumption so first we will convert KJ to J

As we know

1 KJ = 1000 J

36.2 kJ = 36.2 x 1000 = 36200 J

So it means that in chemical reaction 36200 J of heat release for each gram of reactant consumed so how much mass of reactant will be consumed if 1360 J heat will release

Apply unity formula

36200 J of heat release ≅ 1 gram of reactant

1360 J of heat release ≅ X gram of reactant

Do cross multiplication

X gram of reactant = 1 g x 1360 J / 36200 J

X gram of reactant = 0.038 g

So 0.038 g of reactant will produce 1360 J of heat.

5 0

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

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3 years ago

Why don't we use meters or miles when discussion distances between stars or galaxies?

taurus [48]

If we use mile or something else it will be hard to measure in sky .
Because the mile or meters too short for the space,
Galaxy and stars are too long far away from us if we use mile or meter,
It till be take like miliion year to measure.

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2 years ago