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3 years ago

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When 6.36 g of metallic chromium is heated with elemental chlorine gas, 19.37 g of a chromium chloride salt results. Calculate t

he empirical formula of the compound

1 answer:

kupik [55]3 years ago

3 0

Explanation:

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A compound that is composed of molybdenum (Mo) and oxygen (O) was produced in a lab by heating molybdenum over a Bunsen burner.

irakobra [83]

First, we determine the mass of each element from the data collected. We can get the mass of molybdenum Mo from the difference between the mass of crucible and molybdenum and the mass of crucible:
Mass of molybdenum = 39.52 – 38.26 = 1.26 g Mo

We can calculate for the mass of molybdenum oxide from the difference between the mass of crucible and molybdenum oxide and the mass of crucible:
Mass of molybdenum oxide = 39.84 – 38.26 = 1.58g

We can now compute for the mass of oxygen O by subtracting the mass of molybdenum from the mass of molybdenum oxide:
Mass of oxygen in molybdenum oxide = 1.58 – 1.26 = 0.32g O

To convert mass to moles, we use the molar mass of each element.
1.26 g Mo * 1 mol Mo / 95.94 g Mo = 0.0131 mol Mo
0.32 g O * 1 mol O / 15.999 g O = 0.0200 mol O

0.0131 mol is the smallest number of moles. We divide each mole value by this number:
0.0131 mol Mo / 0.0131 = 1
0.0200 mol O / 0.0131 = 1.53

Multiplying these results by 2 to get the lowest whole number ratio,
0.0131 mol Mo / 0.0131 = 1 * 2 = 2
0.0200 mol O / 0.0131 = 1.5 * 2 = 3
Thus, we can write the empirical formula as Mo2O3.

5 0

4 years ago

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In atmospheric chemistry, the following chemical reaction converts SO2, the predominant oxide of sulfur that comes from combusti

Misha Larkins [42]

Answer:

Explanation:

From the given information;

The chemical reaction can be well presented as follows:

$\mathtt{SO_{2(g)} + \dfrac{1}{2}O_{2(g)} }$ ⇄ $\mathtt{3SO_{2(l)}}$

Now, K is known to be the equilibrium constant and it can be represented in terms of each constituent activity:

i.e

$K = \dfrac{a_{so_3}}{a_{so_2} a_{o_2}^{\frac{1}{2}}}$

However, since we are dealing with liquids solutions;

$K = \dfrac{1}{\dfrac{Pso_2}{P^0}\Big ( \dfrac{Po_2}{P^0} \Big)^{1/2}}$ since the activity of $a_{so_3}$ is equivalent to 1

Hence, under standard conditions(i.e at a pressure of 1 bar)

$K = \dfrac{1}{Pso_2Po_2^{1/2}}$

(b)

From the CRC Handbook, we are meant to determine the value of the Gibb free energy by applying the formula:

$\Delta _{rxn} G^o = \sum \Delta_f \ G^o (products) - \sum \Delta_fG^o (reactants) \\ \\ = (1) (-368 \ kJ/mol) - (\dfrac{1}{2}) (0) - ((1) (-300.13 \ kJ/mol)) \\ \\ = -368 \ kJ/mol + 300.13 \ kJ/mol \\ \\ \simeq -68 \ kJ/mol$

Thus, for this reaction; the Gibbs frree energy = -68 kJ/mol

(c)

Le's recall that:

At equilibrium, the instantaneous free energy is usually zero &

Q(reaction quotient) is equivalent to K(equilibrium constant)

So;

$\mathtt{\Delta _{rxn} G = \Delta _{rxn} G^o + RT In Q}$

$\mathtt{0- \Delta _{rxn} G^o = RTIn K } \\ \\ \mathtt{ \Delta _{rxn} G^o = -RTIn K } \\ \\ K = e^{\dfrac{\Delta_{rxn} G^o}{RT}} \\ \\ K = e^{^{\dfrac{67900 \ J/mol}{8.314 \ J/mol \times 298 \ K}} }$

$K =7.98390356\times 10^{11} \\ \\ \mathbf{K = 7.98 \times 10^{11}}$

(d)

The direction by which the reaction will proceed can be determined if we can know the value of Q(reaction quotient).

This is because;

If Q < K, then the reaction will proceed in the right direction towards the products.

However, if Q > K , then the reaction goes to the left direction. i.e to the reactants.

So;

$Q= \dfrac{1}{Pso_2Po_2^{1/2}}$

Since we are dealing with liquids;

$Q= \dfrac{1}{1 \times 1^{1/2}}$

Q = 1

Since Q < K; Then, the reaction proceeds in the right direction.

Hence, SO2 as well O2 will combine to yield SO3, then condensation will take place to form liquid.

8 0

3 years ago

The enthalpies of formation of the compounds in the combustion of methane, , are CH4 (g): Hf = –74.6 kJ/mol; CO2 (g): Hf = –393.

algol [13]

Answer:

The amount of energy released from the combustion of 2 moles of methae is 1,605.08 kJ/mol

Explanation:

The chemical reaction of the combustion of methane is given as follows;

CH₄ (g) + 2O₂ (g) → CO₂ (g) + 2H₂O (g)

Hence, 1 mole of methane combines with 2 moles of oxygen gas to form 1 mole of carbon dioxide and 2 moles of water vapor

Where:

CH₄ (g): Hf = -74.6 kJ/mol

CO₂ (g): Hf = -393.5 kJ/mol

H₂O (g): Hf = -241.82 kJ/mol

Therefore, the combustion of 1 mole of methane releases;

-393.5 kJ/mol × 1 + 241.82 kJ/mol × 2 + 74.6 kJ/mol = -802.54 kJ/mol

Hence the combustion of 2 moles of methae will rellease;

2 × -802.54 kJ/mol or 1,605.08 kJ/mol.

8 0

3 years ago

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I NEED HELP ASAP !!!!!!!!!!!!!!!!!!!!!1

Karolina [17]

Answer:

true im pretttyyy sure

Explanation:

bcuz the stronger the intermolecular forces the higher the boiling point :3

6 0

2 years ago

Fluorine-18 is a positron emitter used in pet scans. write a balanced nuclear equation for the reaction.

kow [346]

<span>Due to limitations on typography, I will have to describe the equation instead of actually writing it. Crude appearance. 18 18 0 F --> O + e 9 8 1 Detailed description. Each of the 3 components have both a left superscript and a left subscript which is a superscript and a subscript to the LEFT of the main figure unlike the usual right side that you see subscripts and superscripts. The equation will be F with an 18 left superscript and a 9 left subscript to represent Florine with atomic weight of 18 and 9 protons. Followed by a right arrow to indicate the direction the reaction is going. Followed by the letter O with a left superscript of 18 and a left subscript of 8 to represent Oxygen with atomic weight of 18 and 8 protons. Followed by a plus sign to indicate more. Followed by either the lower case letter "e" or the upper case Greek character beta with a left superscript of 0 and a left subscript of 1 or +1 to represent the positron being emitted with a positive charge and an atomic weight of 0.</span>

5 0

3 years ago

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