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o-na [289]
3 years ago
8

Choose the thermochemical equation that illustrates ΔH°f for Li2SO4. Choose the thermochemical equation that illustrates ΔH°f fo

r Li2SO4. 8 Li2SO4(s) → 16 Li(s) + S8(s, rhombic) + 16 O2(g) Li2SO4(aq) → 2 Li+(aq) + SO42-(aq) 2 Li+(aq) + SO42-(aq) → Li2SO4(aq) 16 Li(s) + S8(s, rhombic) + 16 O2(g) → 8 Li2SO4(s) 2 Li(s) + 1/8 S8(s, rhombic) + 2 O2(g) → Li2SO4(s)
Chemistry
1 answer:
Deffense [45]3 years ago
8 0

Answer:

2Li(s) + ⅛S₈(s, rhombic) + 2O₂(g) → Li₂SO₄(s)  

Explanation:

A thermochemical equation must show the formation of 1 mol of a substance from its elements in their most stable state,.

The only equation that meets those conditions is the last one.

A and B are wrong , because they show Li₂SO₄ as a reactant, not a product.

C is wrong because Li⁺ and SO₄²⁻ are not elements.

D is wrong because it shows the formation of 8 mol of Li₂SO₄.

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Lice are parasitic insects that live on the scalp, eyebrows, and eyelashes of humans
earnstyle [38]

Answer:

Most likely A

Explanation:

Its a more plausible answer

5 0
3 years ago
The maximum number of electrons in a single d subshell is:
son4ous [18]

10 electrons

Explanation:

The maximum number of electrons in a single d-subshell is 10 electrons.

The d-notation stands for an azimuthal quantum number or secondary quantum number.

This subshell can only accommodate a maximum of 10 electrons.

d- subshell has five orbitals

Each orbital is able to accommodate 2 electrons.

 Therefore 5 x 2 = 10 electrons

learn more:

Atomic orbital brainly.com/question/1832385

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7 0
3 years ago
A student balanced the chemical equation Mg + O2 →MgO by writing Mg + O2 → MgO2. Was the equation balanced correctly? Explain yo
Amanda [17]

Explanation:

Charges on both magnesium and oxygen is 2. Though opposite in sign, they have equal charges so, both of them will be cancelled by each other.

As a result, formula of magnesium oxide is MgO and not MgO_{2}.

The student write the equation as Mg + O_2 \rightarrow MgO_2, it is not correct.

Therefore, given equation will be balanced as follows.

          2Mg + O_{2} \rightarrow 2MgO

Since, number of atoms on both reactant and product side are equal. Hence, this equation is completely balanced.

7 0
3 years ago
The osmotic pressure of a solution containing 2.04 g of an unknown compound dissolved in 175.0 mLof solution at 25 ∘C is 2.13 at
kherson [118]

<u>Answer:</u> The molecular formula of the compound is C_4H_{10}O_4

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=iMRT

Or,

\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT

where,

\pi = osmotic pressure of the solution = 2.13 atm

i = Van't hoff factor = 1 (for non-electrolytes)

Given mass of compound = 2.04 g

Volume of solution = 175.0 mL

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature of the solution = 25^oC=[273+25]=298K

Putting values in above equation, we get:

2.13atm=1\times \frac{2.04\times 1000}{\text{Molar mass of compound}\times 175.0}\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\\text{Molar mass of compound}=\frac{1\times 2.04\times 1000\times 0.0821\times 298}{2.13\times 175.0}=133.9g/mol

  • <u>Calculating the molecular formula:</u>

The chemical equation for the combustion of compound having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=36.26g

Mass of H_2O=14.85g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 36.26 g of carbon dioxide, \frac{12}{44}\times 36.26=9.89g of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18 g of water, 2 g of hydrogen is contained.

So, in 14.85 g of water, \frac{2}{18}\times 14.85=1.65g of hydrogen will be contained.

Mass of oxygen in the compound = (22.08) - (9.89 + 1.65) = 10.54 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = \frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{9.89g}{12g/mole}=0.824moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.65g}{1g/mole}=1.65moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{10.54g}{16g/mole}=0.659moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.659 moles.

For Carbon = \frac{0.824}{0.659}=1.25\approx 1

For Hydrogen = \frac{1.65}{0.659}=2.5

For Oxygen = \frac{0.659}{0.659}=1

Converting the mole fraction into whole number by multiplying the mole fraction by '2'

Mole fraction of carbon = (1 × 2) = 2

Mole fraction of oxygen = (2.5 × 2) = 5

Mole fraction of hydrogen = (1 × 2) = 2

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 5 : 2

The empirical formula for the given compound is C_2H_5O_2

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

n=\frac{\text{Molecular mass}}{\text{Empirical mass}}

We are given:

Mass of molecular formula = 133.9 g/mol

Mass of empirical formula = 61 g/mol

Putting values in above equation, we get:

n=\frac{133.9g/mol}{61g/mol}=2

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(2\times 2)}H_{(5\times 2)}O_{(2\times 2)}=C_4H_{10}O_4

Hence, the molecular formula of the compound is C_4H_{10}O_4

4 0
3 years ago
What happens if the metal you throw in is MORE REACTIVE than the<br> metal ion in solution?
IRISSAK [1]
A displacement reaction Would occur in this situation
6 0
3 years ago
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