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3 years ago

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The thermite reaction, in which powdered aluminum reacts with iron oxide, is highly exothermic: 2al(s) + fe2o3(s)→al2o3(s) + 2fe

(s) use standard enthalpies of formation to find δh∘rxn for the thermite reaction. express the heat of the reaction in kilojoules to four significant figures.

1 answer:

ycow [4]3 years ago

7 0

Answer is: <span>heat of the reaction is </span>-851,500 kJ.
Chemical reaction: 2Al + Fe₂O₃ → Al₂O₃ + 2Fe.
ΔHf(Fe₂O₃) = -824,2 kJ/mol; enthalpy of formation.
ΔHf(Al₂O₃) = -1675,7 kJ/mol.
By definition ΔH(element) = 0 kJ/mol.
ΔHrxn = ∑ΔH(products) - ∑ΔH(reactants).
ΔHrxn = 1 mol · (-1675,7 kJ/mol) - 1 mol · (-824,2 kJ/mol).
ΔHrxn = -851,500 kJ.

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Suppose a group of volunteers is planning to build a park near a local lake. The lake is known to contain low levels of arsenic

Kisachek [45]

Answer:

A) 10.75 is the concentration of arsenic in the sample in parts per billion .

B) 7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C) It will take 1.37 years to remove all of the arsenic from the lake.

Explanation:

A) Mass of arsenic in lake water sample = 164.5 ng

The ppb is the amount of solute (in micrograms) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation:

$\text{ppb}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^9$

Both the masses are in grams.

We are given:

Mass of arsenic = 164.5 ng = $164.5\times 10^{-9} g$

$1 ng=10^{-9} g$

Volume of the sample = V = $15.3 cm^3$

Density of the lake water sample ,d= $1.00 g/cm^3$

Mass of sample = M = $d\times V=1.0 g/cm^3\times 15.3 cm^3=15.3 g$

$ppb=\frac{164.5\times 10^{-9} g}{15.3 g}\times 10^9=10.75$

10.75 is the concentration of arsenic in the sample in parts per billion.

B)

Mass of arsenic in $1 cm^3$ of lake water = $\frac{164.5\times 10^{-9} g}{15.3}=1.075\times 10^{-8} g$

Mass of arsenic in $0.710 km^3$ lake water be m.

$1 km^3=10^{15} cm^3$

Mass of arsenic in $0.710\times 10^{15} cm^3$ lake water :

$m=0.710\times 10^{15}\times 1.075\times 10^{-8} g=7,633,660.130 g$

1 g = 0.001 kg

7,633,660.130 g = 7,633,660.130 × 0.001 kg=7,633.660130 kg ≈ 7,633.66 kg

7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C)

Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.

Days required to remove 1 kilogram of arsenic from the lake water :

$\frac{2.74}{41.90} days$

Then days required to remove 7,633.66 kg of arsenic from the lake water :

$=7,633.66\times \frac{2.74}{41.90} days=499.19 days$

1 year = 365 days

499.19 days = $\frac{499.19}{365} years = 1.367 years\approx 1.37 years$

It will take 1.37 years to remove all of the arsenic from the lake.

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3 years ago

The scientific theory is called

Aliun [14]

Answer:

is a well-sustantiated explanation of some aspect of the natural world, based on a body of facts that have been repeatedly confirmed through observation and experiment.such fact- supported theories are not "guesses" but reliable accounts of the real worlds.

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3 years ago

The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +

Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) The solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

Explanation:

(a) Chemical equation for the given reaction in pure water is as follows.

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$

Initial: 0 0

Change: +x +x

Equilibm: x x

$K_{sp} = 1.2 \times 10^{-6}$

And, equilibrium expression is as follows.

$K_{sp} = [Cu^{+}][Cl^{-}]$

$1.2 \times 10^{-6} = x \times x$

x = $1.1 \times 10^{-3} M$

Hence, the solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) When NaCl is 0.1 M,

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$ , $K_{sp} = 1.2 \times 10^{-6}$

$Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq)$ , $K = 8.7 \times 10^{4}$

Net equation: $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

$K' = K_{sp} \times K$

= 0.1044

So for, $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

Initial: 0.1 0

Change: -x +x

Equilibm: 0.1 - x x

Now, the equilibrium expression is as follows.

K' = $\frac{CuCl_{2}}{Cl^{-}}$

0.1044 = $\frac{x}{0.1 - x}$

x = $9.5 \times 10^{-3} M$

Therefore, the solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

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Are Si valence electrons shared? <br><br> Are C valence electrons shared?

MariettaO [177]

Answer:

4

Explanation:

Si has atomic number 14 so the electronic configuration 2,8,4

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Calcium solid is placed in a solution of iron (III) chloride producing calcium chloride and iron solid

mestny [16]

Answer:

3Ca₍s₎ + 2FeCl₃ -------------------------------------------------------------> 3CaCl₂ + 2Fe₍s₎.

Explanation:

Iron(III) chloride is also known as Iron trichloride and it has the molar mass value of 162.2 g/mol and density of 2.9 g/cm³.

<em>The balanced equation for the chemical reaction between solid calcium and iron (III) chloride is given below as: </em>

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3Ca₍s₎ + 2FeCl₃ -------------------------------------------------------------> 3CaCl₂ + 2Fe₍s₎.

This kind of chemical reaction is known as <em>single displacement reaction or single replacement reaction. </em>

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