3. Mercury is a liquid metal that has a density of 13.58 g/mL. Calculate the volume of mercury

What is the empirical formula of a compound containing 5.03 grams carbon, 0.42 grams hydrogen, and 44.5 grams chlorine?

Thepotemich [5.8K]

Answer:

CHCl₃

Explanation:

Given parameters:

Carbon = 5.03g

Hydrogen = 0.42g

Chlorine = 44.5g

The empirical formula shows the simplest formula of a compound.

To deduce the empirical, we need two pieces of information:

> Mass of the elements or the percentage composition of the compound

>The relative atomic masses of the elements

In order to derive the empirical formula from these parameters,

>>> find the number of moles of elements by dividing the mass given by the relative atomic mass of the respective atom

>>> Divide through by the smallest mole

>>> Approximate or multiply by a factor that would make it possible for whole numbers to be obtained

From the question, we have been given the mass of each element.

Now using the period table, we can obtain the relative atomic masses of each atom:

Carbon = 12gmol⁻¹

Hydrogen = 1gmol⁻¹

Chlorine = 37.5gmol⁻¹

C H Cl

Mass(in g) 5.03 0.42 44.5

Moles 5.03/12 0.42/1 44.5/37.5

0.42 0.42 1.19

Dividing

by

smallest 0.42/0.42 0.42/0.42 1.19/0.42

Mole ratio 1 1 2.83

Approximate 1 1 3

The empirical formula is CHCl₃

8 0

2 years ago

The rate of disappearance of HCl was measured for the following reaction:

AURORKA [14]

Answer: (a) 0.000083M/s, 0.000069M/s, 0.000052M/s, 0.000034M/s

(b) average reaction rate between t=0.0min to t=430.0min =0.000049M/s

(c) The average rate between t=54.0 and t=215.0min is greater than the average rate between t=107.0 and t=430.0min

Explanation:

Average reaction rate = change in concentration / time taken

(a) after 54mins, t = 54*60s = 3240s

average reaction rate = (1.58 - 1.85)M / (3240 * 0.0)s

= -0.27M/3240

= 0.000083M/s

after 107mins, t = 107*60s = 6420s

average reaction rate = (1.36 - 1.58)M/ (6420 - 3240)s

= -0.22M/3180s

= 0.000069M/s

after 215mins, t = 215*60s = 12900s

average reaction rate = (1.02 - 1.36)M/ (12900 - 6420)s

= -0.34M/6480s

= 0.000052M/s

after 430mins,t = 430*60 = 25800s

average reaction rate = (0.580 - 1.02)M / (25800 - 12900)s

= -0.44M/12900s

= 0.000034M/s

(b) average reaction rate between t=0.0min to t=430.0min

= (0.580 - 1.85)M/ (25800 - 0.0)s

= -1.27M/25800s

=0.000049M/s

(c) average reaction rate between t = 54.0min and t = 215.0min

= (1.02 - 1.58)M / (12900 - 3240)s

= -0.56M/9660s

= 0.000058M/s

average reaction rate between t=107.0 and t=430.0min

= (0.580 - 1.36)M / (25800 - 6420)s

= 0.78M /19380s

= 0.000040M/s

Therefore the average rate between t=54.0 and t=215.0min is greater than the average rate between t=107.0 and t=430.0min

6 0

2 years ago

Which of these do not obey the octet rule? select all that apply. select all that apply. clo clo− clo2− clo3− clo4−?

ioda

The correct answer is ClO, ClO3-, ClO- and ClO4-

Kossel and Lewis in 1916 developed an important theory of chemical combination between atoms known as electronic theory of chemical bonding. According to this, atoms can combine either by transfer of valence electrons from one atom to another (gaining or losing) or by sharing of valence electron in order to have an octet( 8 electron) in their shells. This is known as octet rule.

In ClO2-, oxygen contains 8 electrons in its valence shell and oxygen will share one electron with chlorine to complete the octet of Cl. In other four, we can clearly see that there are more or less than 8 electrons in the outer shell of oxygen so we can clearly say that ClO, ClO3-, ClO- and ClO4- are disobeying the octet rule.

3 0

2 years ago

Read 2 more answers

The amount of gas that occupies 36.52 L at 68.0°C and 672 mm Hg is __________ mol.

ASHA 777 [7]

We assume that this gas is an ideal gas. We use the ideal gas equation to calculate the amount of the gas in moles. It is expressed as:

PV = nRT
(672) (1/760) (36.52) = n (0.08206) ( 68 +273.15)
n = 1.15 mol of gas

Hope this answers the question. Have a nice day.

7 0

1 year ago

A bottle of concentrated aqueous sulfuric acid, labeled 98.0 wt% h2so4, has a concentration of 18.0 m. (a) how many milliliters

nadya68 [22]

n this order, Ď=1.8gmL, cm=0.5, and mole fraction = 0.9 First, let's start with wt%, which is the symbol for weight percent. 98wt% means that for every 100g of solution, 98g represent sulphuric acid, H2SO4. We know that 1dm3=1L, so H2SO4's molarity is C=nV=18.0moles1.0L=18M In order to determine sulphuric acid solution's density, we need to find its mass; H2SO4's molar mass is 98.0gmol, so 18.0moles1Lâ‹…98.0g1mole=1764g1L Since we've determined that we have 1764g of H2SO4 in 1L, we'll use the wt% to determine the mass of the solution 98.0wt%=98g.H2SO4100.0g.solution=1764gmasssolutionâ†’ masssolution=1764gâ‹…100.0g98g=1800g Therefore, 1L of 98wt% H2SO4 solution will have a density of Ď=mV=1800g1.0â‹…103mL=1.8gmL H2SO4's molality, which is defined as the number of moles of solute divided by the mass in kg of the solvent; assuming the solvent is water, this will turn out to be cm=nH2SO4masssolvent=18moles(1800â’1764)â‹…10â’3kg=0.5m Since mole fraction is defined as the number of moles of one substance divided by the total number of moles in the solution, and knowing the water's molar mass is 18gmol, we could determine that 100g.solutionâ‹…98g100gâ‹…1mole98g=1 mole H2SO4 100g.solutionâ‹…(100â’98)g100gâ‹…1mole18g=0.11 moles H2O So, H2SO4's mole fraction is molefractionH2SO4=11+0.11=0.9

5 0

1 year ago