All categories

3 years ago

Calculate the empirical formula for a compound containing 67.6% Hg 10.8% S 21.6% O

Chemistry

1 answer:

pashok25 [27]3 years ago

3 0

Answer:

HgSO₄

Explanation:

% => g => moles => ratio => reduce => empirical ratio

%Hg = 67.6% => 67.6g/201g/mol = 0.34mol

%S = 10.8% => 10.8g/32g/mol = 0.34mol

%O = 21.6% => 21.6g/16g/mol = 1.35mol

Hg:S:O => 0.34:0.34:1.35

Reduce to whole number ratio by dividing by the smaller mole value...

Hg:S:O => 0.34/.34:0.34/.34:1.35/.34 => Empirical Ratio = 1:1:4

∴ Empirical Formula is HgSO₄

You might be interested in

Identify the reactants and products in the following chemical equation: Zn +2HCl --> ZnCl2 + H2 *

Ad libitum [116K]

Answer:

Explanation:

zinc and hydrochloric acid reacts to form zinc chloride and hydrogen.

hope it helps. :)

7 0

2 years ago

If a solid has a density of 4.0 g/cm^3, what volume of the solid has a mass<br> of 24 g? Show work.

Mrrafil [7]

Answer:

<h2>Volume = 6 cm³</h2>

Explanation:

Density of a substance is given by

$Density = \frac{mass}{volume}$

From the question

Density = 4.0 g/cm³

mass = 24g

Substitute the values into the above formula and solve for the volume

That's

$4 = \frac{24}{v} \\4v = 24$

Divide both sides by 4

v = 6

We have the final answer as

<h3>Volume = 6 cm³</h3>

Hope this helps you

8 0

3 years ago

2. Electron A falls from energy level X to energy level Y and releases blue light. Electron B falls from energy level Y to energ

Lynna [10]

Wouldnt you find that on the grid or the table

3 0

3 years ago

Read 2 more answers

How many moles are in 297 g of NH3?

BigorU [14]

Answer:

1. 17.4 moles.

2. 1.13 moles

3. 315.5 moles

4. 390.6g

5. 1.13x10⁶ moles

6. 14.8 moles

7. 337 moles

8. 2.15x10²⁴ molecules

9. 3.13x10²⁴ atoms

10. 1.38x10²⁴ particles

11. 1517g

12. 455g of CaF₂

Explanation:

We can convert formula units to moles or vice versa using Avogadro's number and moles to grams using molar mass of the substance:

1. Molar mass NH3: 17.031g/mol

297g * (1mol / 17.031g) = 17.4 moles

2. Molar mass MgCO3: 84.3g/mol

95g * (1mol / 84.3g) = 1.13 moles

3. Using Avogadro's number (6.022x10²³ formula units / mol):

1.9x10²⁶FU * (1mol / 6.022x10²³FU) = 315.5 moles

4. Molar mass H2O: 18g/mol

21.7mol * (18g / mol) = 390.6g

5. Using Avogadro's number (6.022x10²³ molecules / mol):

6.78x10²⁹molecules * (1mol / 6.022x10²³FU) = 1.13x10⁶ moles

6. 8.9x10²⁴FU * (1mol / 6.022x10²³FU) = 14.8 moles

7. Using Avogadro's number (6.022x10²³ atoms / mol):

2.03x10²⁶atoms* (1mol / 6.022x10²³FU) = 337 moles

8. 3.569mol * (6.022x10²³ molecules / 1mol) = 2.15x10²⁴ molecules

9. 5.2mol * (6.022x10²³ atoms / 1mol) = 3.13x10²⁴ atoms

10. Molar mass Li₂SO₄: 109.94g/mol:

36g * (1mol / 109.94g) * (6.022x10²³ molecules / 1mol) * (7 particles / 1molecule) = 1.38x10²⁴ particles

<em>Assuming particles are atoms and in 1 molecule of Li₂SO₄ you have 7 atoms.</em>

11. Molar mass Cl₂: 70.9g/mol:

21.4mol * (70.9g / mol) = 1517g

12. Molar mass CaF₂: 78.07g/mol:

3.51x10²⁴FU * (1mol / 6.022x10²³FU) * (78.07g / mol) = 455g of CaF₂

8 0

3 years ago

When 0.250 moles of KCl are added to 200.0 g of water in a constant pressure calorimeter a temperature change is observed. Given

777dan777 [17]

Explanation:

Upon dissolution of KCl heat is generated and temperature of the solution raises.

Therefore, heat generated by dissolving 0.25 moles of KCl will be as follows.

$17.24 kJ/mol \times 0.25 mol$

= 4.31 kJ

or, = 4310 J (as 1 kJ = 1000 J)

Mass of solution will be the sum of mass of water and mass of KCl.

Mass of Solution = mass of water + (no. of moles of KCl × molar mass)

= 200 g + $(0.25 mol \times 54.5 g/mol)$

= 200 g + 13.625 g

= 213.625 g

Relation between heat, mass and change in temperature is as follows.

Q = $mC \Delta T$

where, C = specific heat of water = $4.184 J/g^{o}C$

Therefore, putting the given values into the above formula as follows.

Q = $mC \Delta T$

4310 J = $213.625 g \times 4.184 J/g^{o}C \times \Delta T$

$\Delta T = 4.82^{o}C$

Thus, we can conclude that rise in temperature will be $4.82^{o}C$ .

6 0

3 years ago