The same chemistry student has a weight of 155lbs what is the student weight in grams

A compound is analyzed and found to contain 22.10%Al, 25.40%P, and 52.50%O. What is the empirical formula of the compound?

Sergio039 [100]

We will determine the moles of each element first

moles = mass / atomic mass

Atomic weight of Al : 27 g/ mol

Atomic weight of P : 3 1g /mol

Atomic weight of O : 16 g /mol

Let the total mass of compound is 100g

The mass of each element will be

Al = 22.10 g

P = 25.40 g

O = 52.50 g

Moles of Al = mass / atomic mass = 22.10 / 27 = 0.819

Moles of P = mass / atomic mass = 25.40/ 31 = 0.819

Moles of O = mass / atomic mass = 52.50/ 16 = 3.28

Now we will divide the moles of each element with the lowest moles obtained to obtain a whole number ratio of moles of each element present

moles of Al = 0.819 / 0.819 = 1

moles of P = 0.819 / 0.819 = 1

moles of O = 3.28 / 0.819 = 4

So the empirical formula will be : AlPO4

7 0

4 years ago

How many L of carbon dioxide at 1.00 atm and 298.15 K are released from a car's engine upon consumption of a 60.0 L LIQUID tank

torisob [31]

Answer:

79,000 liters

Explanation:

1. Number of moles of gasoline

a) Convert 60.0 liters to grams

density = 0.77kg/liter

density = mass / volume

mass = density × volume

mass = 0.77kg/liter × 60.0 liter = 46.2 kg

46.2kg × 1,000g/kg = 46,200g

b) Convert 46,200 grams to moles

molar mass of C₈H₁₈ = 114.2 g/mol

number of moles = mass in grams / molar mass

number of moles = 46,200g / (114.2 gmol) = 404.55 mol

2. Number of moles of carbon dioxide, CO₂ produced

a) Balanced chemical equation (given):

C₈H₁₈ (l) + ²⁵/₂ O₂ (g) → 8 CO₂ (g) + 9 H₂O (g)

b) mole ratio:

1 mol C₈H₁₈ / 8 mol CO₂ = 404.55 mol C₈H₁₈ / x

Solve for x:

x = 404.55mol C₈H₁₈ × 8 mol CO₂ / 1mol C₈H₁₈ = 3,236.4 mol CO₂

3. Convert the number of moles of carbon dioxide to volume

Use the ideal gas equation:

pV = nRT

V = nRT/p

p = 1 atm

T = 298.15K

n = 3,236.4 mol

R = 0.08206 (mol . liter)/ (K . mol)

Substitute and compute:

V =3,236.4 mol × 0.08206 (mol . liter) / (K . mol) 298.15K / 1 atm

V = 79,183 liter

Round to two significant figures (because the density has two significant figures): 79,000 liters ← answer

3 0

3 years ago

Determine the point groups for a. Naphthalene b. 1,8-Dichloronaphthalene c. 1,5-Dichloronaphthalene d. 1,2-Dichloronaphthalene

Sloan [31]

Answer:

a). Nepthalene is a D2h molecule because it has 3 C2 axes which are perpendicular and it has a mirror plane which is horizontal. Moreover C2 is taken as principal axis.

b). 1,8-dichloronaphthalene is a C2v molecule beccause it has 1 C2 axis, two rings are joined by the C-C bond and it also has two mirro planes.

c). 1,5-dichloronaphthalene is a C2h molecule because it has only 1 C2 axis which is pependicular to the plane, it has an inversion center and also a mirror plane which is horizontal in position.

d). 1,2-dichloronaphthalene is a Cs molecule because it has only a mirror plane.

7 0

4 years ago

Linolenic acid (C18H30O2 - M.W. = 278.42 g/mol) reacts with hydrogen gas according to the equation: C18H30O2 + 3H2 (g) → C18H36O

Ludmilka [50]

Answer:

2.53 L is the volume of H₂ needed

Explanation:

The reaction is: C₁₈H₃₀O₂ + 3H₂ → C₁₈H₃₆O₂

By the way we can say, that 1 mol of linolenic acid reacts with 3 moles of oxygen in order to produce, 1 mol of stearic acid.

By stoichiometry, ratio is 1:3

Let's convert the mass of the linolenic acid to moles:

10.5 g . 1 mol / 278.42 g = 0.0377 moles

We apply a rule of three:

1 mol of linolenic acid needs 3 moles of H₂ to react

Then, 0.0377 moles will react with (0.0377 . 3 )/1 = 0.113 moles of hydrogen

We apply the Ideal Gases Law to find out the volume (condition of measure are STP) → P . V = n . R . T → V = ( n . R .T ) / P

V = (0.113 mol . 0.082 L.atm/mol.K . 273.15K) 1 atm = 2.53 L

4 0

3 years ago

PLEASE HELP: Calculate the average atomic mass of the following elements given the information below (show work): Hydrogen, whic

k0ka [10]

99.98%=0.9998
0.01%=0.0001
=(1)(0.9998)+2(0.0001)+3(0.0001)
=1.0003

3 0

3 years ago