1. How many moles of O, are needed to burn 58.7g of propane (C2H3) according to the

Alisiya [41]

Answer:

3.98 C my friend you welcome

6 0

3 years ago

The densities of gases a, b, and c at stp are 1.25 g/l, 2.86 g/l and 0.714 g/l, respectively. calculate the molar mass of each s

kicyunya [14]

The molar mass of a, b and c at STP is calculated as below

At STP T is always= 273 Kelvin and ,P= 1.0 atm

by use of ideal gas equation that is PV =nRT
n(number of moles) = mass/molar mass therefore replace n in the ideal gas equation

that is Pv = (mass/molar mass)RT
multiply both side by molar mass and then divide by Pv to make molar mass the subject of the formula

that is molar mass = (mass x RT)/ PV

density is always = mass/volume

therefore by replacing mass/volume in the equation by density the equation
molar mass=( density xRT)/P where R = 0.082 L.atm/mol.K

the molar mass for a
= (1.25 g/l x0.082 L.atm/mol.k x273k)/1.0atm = 28g/mol

the molar mass of b
=(2.86g/l x0.082L.atm/mol.k x273 k) /1.0 atm = 64 g/mol

the molar mass of c

=0.714g/l x0.082 L.atm/mol.K x273 K) 1.0atm= 16 g/mol

therefore the
gas a is nitrogen N2 since 14 x2= 28 g/mol
gas b =SO2 since 32 +(16x2)= 64g/mol
gas c = methaneCH4 since 12+(1x4) = 16 g/mol

8 0

3 years ago

A gas has a pressure of 8.5atm and occupies 24L at 25∘C. What volume (in liters) will the gas occupy if the pressure is increase

Vesna [10]

The volume (in liters) that the gas will occupy if the pressure is increased to 13.5 atm and the temperature is decreased to 15 °C is 15 L

From the question given above, the following data were obtained:

Initial pressure (P₁) = 8.5 atm

Initial volume (V₁) = 24 L

Initial temperature (T₁) = 25 °C = 25 + 273 = 298 K

Final pressure (P₂) = 13.5 atm

Final temperature (T₂) = 15 °C = 15 + 273 = 288 K

<h3>Final volume (V₂) =? </h3>

The final volume of the gas can be obtained by using the combined gas equation as illustrated below:

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\\\frac{8.5 * 24}{298} = \frac{13.5 * V_{2}}{288}\\\\ \frac{204}{298} = \frac{13.5 * V_{2}}{288}\\\\$

Cross multiply

298 × 13.5 × V₂ = 204 × 288

4023 × V₂ = 58752

Divide both side by 4023

$V_{2} = \frac{58752}{4023}\\\\$

Therefore, the final volume of the gas is 15 L

Learn more: brainly.com/question/25547148

3 0

2 years ago

The quantum numbers for the last electron placed in three elements are listed below. Which of these is(are) NOT correct? Er (4 3

UNO [17]

Answer:

The three elements Erbium, Thallium and Osmium have incorrect quantum numbers for the last electron placed.

Explanation:

The 4 quantum numbers are (n,l,ml,ms):

n (Principal quantum number): it is the number of the shell (level) where the electron is placed.
l (Angular momentum quantum number or Secondary): it represents the sublevel where the electron is placed. There are 4 subleves: s, p d and f so secondary quantum number can take the number 0 (s), 1 (p), 2 (d) or 3 (f) depending on which sublevel the electron is placed.
ml (Magnetic quantum number): it represents the spatial orientation of the electron in respect of the sublevel the electron is placed. For example: if the electron occupies the s sublevel the magnetic number will be 0, if the electron occupies the p sublevel the magnetic number could be -1,0,1, if the electron occupies the d sublevel the magnetic number could be -2,-1,0,1,2 and if the electron occupies the f sublevel the magnetic number could be -3,-2,-1,0,1,2,3. You can see this in the attachment related to the correct sublevel for the example.

ms (Spin quantum number): this number represents the possible rotation of the electron so it could be 1/2 (which is represented by an up arrow) or -1/2 (represented by an down arrow).

Let's analyze the last electron of each element. You can see the attachment for better understanding. The last electron it is represented with orange color.

- Erbium:

This element has 68 electrons so following the Moeller's Diagram to fill the the electronic configuration, we found that the last electron of Erbium it is in the 4th level (shell), in the f sublevel. As Erbium has 12 electrons in the f sublevel, it is necessary to follow the Hund's rule (electrons must be placed singly in every sublevel before place a parallel electron) to placed correctly all of them. Finally, the last electron of Erbium stays in the middle of the sublevel and it is represented by a down arrow so the correct quantum numbers in the Erbium element are (4,3,1,-1/2).

- Thallium:

This element has 81 electrons and following the Moeller's Diagram, we found that it last electron it is in the 6th level, in the p sublevel. As Thallium has 1 electron in the p sublevel, it is placed singly in the sublevel. So the last electron of Thallium it is represented by an up arrow so the correct quantum numbers in the Thallium element are (6,1,-1,1/2).

- Osmium:

Osmium has 76 electrons and following the steps that we did with we the other elements, we noticed that its last electron it is in the 5th level, in the d sublevel. Following the Hund's rule the last electron of Osmium has a magnetic quantum number of -2 and its spin quantum number is -1/2, so the quantum numbers in the Osmium element are (5,2,-2,-1/2).

Note:

- Remember that the s sublevel has place for 2 electrons, the p sublevel has place for 6 electrons, the d sublevel has place for 10 electrons and the f sublevel has place for 14 electrons.

3 0

3 years ago

How come flowers usually contain more stamen the pistils

USPshnik [31]

To make sure there are enough pollen to fertilize the seeds from the pistil. Not all pollen can make it to the pistil. It's the same as animals producing millions of sperms although only one is needed for fertilization.

5 0

3 years ago