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3 years ago

Calculate the amount of energy released by the freezing of 13.3 g of water

Chemistry

1 answer:

timama [110]3 years ago

4 0

We can calculate for the amount of heat released using the formula:

ΔH = ʎ m

where,

ΔH = heat released

ʎ = latent heat of fusing for water = - 333.55 J/g

m = mass of water = 13.3 g

Calculating:

ΔH = (- 333.55 J/g) * 13.3 g

ΔH = -4,503 J = 1.08 cal

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Andreas93 [3]

Answer:

Explanation:

bc inorganic compoud refers to all compound that do not contain carbons.

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3 years ago

List at least two chemical reactions that resulted in a yellow, orange, or red precipitate. For these reactions, list the possib

Katarina [22]

<span>1) 0.2M ferric nitrate is added gradually to 1M sodium hydroxide. In result, a red precipitate appears. The precipitate is ferric hydroxide.
2) </span><span>0.2M potassium chromate is added gradually to 0.05M lead acetate. in result, a yellow precipitate appears. The precipitate is called potassium acetate.
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3 years ago

The following reactions can be used to prepare samples of metals. Determine the enthalpy change under standard state conditions

mamaluj [8]

Answer:

a) 62.1 kJ/mol

b) 2.82 kJ/mol

c) 270.91 kJ/mol

d) -851.5 kJ/mol

Explanation:

The enthalpy change for a reaction in standard conditions (ΔH°rxn) can be calculated by:

ΔH°rxn = ∑n*ΔH°f, products - ∑n*ΔH°f, reagents

Where n is the number of moles in the stoichiometry reaction, and ΔH°f is the enthalpy of formation at standard conditions. ΔH°f = 0 for substances formed by only a single element. The values can be found in thermodynamics tables.

a) 2Ag₂O(s) → 4Ag(s) + O₂(g)

ΔH°f, Ag₂O(s) = -31.05 kJ/mol

ΔH°rxn = 0 - (2*(-31.05)) = 62.1 kJ/mol

b) SnO(s) + CO(g) → Sn(s) + CO₂(g)

ΔH°f,SnO(s) = -285.8 kJ/mol

ΔH°f,CO(g) = -110.53 kJ/mol

ΔH°f,CO₂(g) = -393.51 kJ/mol

ΔH°rxn = [-393.51] - [-110.53 - 285.8] = 2.82 kJ/mol

c) Cr₂O₃(s) + 3H₂(g) → 2Cr(s) + 3H₂O(l)

ΔH°f,Cr₂O₃(s) = -1128.4 kJ/mol

ΔH°f,H₂O(l) = -285.83 kJ/mol

ΔH°rxn = [3*(-285.83)] - [( -1128.4)] = 270.91 kJ/mol

d) 2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(s)

ΔH°f,Fe₂O₃(s) = -824.2 kJ/mol

ΔH°f,Al₂O₃(s) = -1675.7 kJ/mol

ΔH°rxn = [-1675.7] - [-824.2] = -851.5 kJ/mol

3 0

3 years ago

a compound with the empirical formula CH2 has a molar mass of 112 g/mol. What is the molecular formula for this compound?

miv72 [106K]

Hope it cleared your doubt :-D

5 0

3 years ago

During a chemical reacion, an iron atom beacme the ion Fe+2. what happened to the iron atom?

galben [10]

Answer:

That iron atom is oxidized. It loses two electrons.

Explanation:

Compare the formula of an iron atom and an iron(II) ion:

Iron atom: $\mathrm{Fe}$ ;
Iron(II) ion: $\mathrm{Fe^{2+}}$ .

The superscript $+2$ in the iron(II) ion is the only difference between the two formulas. This superscript indicates a charge of $+2$ on each ion. Atoms and ions contain protons. In many cases, they also contain electrons. Each proton carries a positive charge of $+1$ and each electron carries a charge of $-1$ . Atoms are neutral for they contain an equal number of protons and electrons.

Protons are located at the center of atoms inside the nuclei. They cannot be gained or lost in chemical reactions. However, electrons are outside the nuclei and can be gained or lost. When an atom loses one or more electrons, it will carry more positive charge than negative charge. It will becomes a positive ion. Conversely, when an atom gains one or more electrons, it becomes a negative ion.

An iron atom $\mathrm{Fe}$ will need to lose two electrons to become a positive iron(II) ion $\mathrm{Fe^{2+}}$ with a charge of $+2$ on each ion. That is:

$\rm Fe \to Fe^{2+} + 2\;e^{-}$ .

Oxidation is Losing one or more electrons;
Reduction is Gaining one or more electrons.

This definition can be written as the acronym OILRIG. (Khan Academy.)

In this case, each iron atom loses two electrons. Therefore the iron atoms here are oxidized.

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3 years ago