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3 years ago

Basic Chem:

Chemistry

1 answer:

Agata [3.3K]3 years ago

5 0

1. 2Fes2 + 5.5O2. Fe2O3 + 4SO2

2. is already balanced

3. 2k + Br2. KBr

4. SiO2 + 4HF. SiF4 + 2H2O

hope that helps

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Which situation would be considered pseudoscience?

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The lucky necklace as it can't be proven.

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A 250 cm^3 solution containing 1,46 g of sodium chloride is added to an excess of silver nitrate solution. The reaction is given

faust18 [17]

Answer:

The mass of the precipitate that AgCl is 3.5803 g.

Explanation:

a) To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Mass of solute (NaCl) = 1.46 g

Molar mass of sulfuric acid = 58.5 g/mol

Volume of solution = $250 cm^3 =250 mL$

$1 cm^3= 1 ml$

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{1.46g\times 1000}{58.5g/mol\times 250}\\\\\text{Molarity of solution}=0.09982 M$

0.09982 M is the concentration of the sodium chloride solution.

b) $NaCl (aq)+AgNO_3 (aq)\rightarrow AgCI(s)+NaNO_3(aq)$

Moles of NaCl = $\frac{1.46 g}{58.5 g/mol}=0.02495 mol$

according to reaction 1 mol of NaCl gives 1 mol of AgCl.

Then 0.02495 moles of NaCl will give:

$\frac{1}{1}\times 0.02495 mol=0.02495 mol$ of AgCl

Mass of 0.02495 moles of AgCl:

$0.02495 mol\times 143.5 g/mol=3.5803 g$

The mass of the precipitate that AgCl is 3.5803 g.

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3 years ago

Categorize them by Metal, Nonmetal, Metalloid

laila [671]

Answer:

METAL: found in periodic table, lithium, shiny, lose electrons easily, good conductor, elements

NONMETAL: brittle, ductile, semimetals, found in periodic table, often gain electrons, semiconductors, carbon, shiny, poor conductor, elements

METALLOID: solid, non- ductile, malleable, found in periodic table, silicon, shiny, can be liquids, elements

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

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