answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Masja [62]
1 year ago
13

Assuming ideal behavior, how many moles of argon would you need to fill a 14.0×12.0×10.0 ft room? assume atmospheric pressure of

1.00 atm, a room temperature of 20.0 ∘c, and 28.2 l/ft3. express your answer with the appropriate units.
Chemistry
1 answer:
Blababa [14]1 year ago
4 0

We use the formula:

PV = nRT

First let us get the volume V:

volume = 14 ft * 12 ft * 10 ft = 1,680 ft^3

Convert this to m^3:

volume = 1680 ft^3 * (1 m / 3.28 ft)^3 = 47.61 m^3

 

n = PV / RT

n = (1 atm) (47.61 m^3) / (293.15 K * 8.21x10^-5 m3 atm / mol K)

<span>n = 1,978.13 mol</span>

You might be interested in
How many structures are possible for a trigonal bipyramidal molecule with a formula of ax3y2?
ryzh [129]
Check the attached file for the answer.

4 0
1 year ago
Consider a beaker of water sitting on the pan of an electronic scale that has been tared. A metal weight hanging from a string i
natka813 [3]

Answer:

See explanation below for answers

Explanation:

We know that the balance is tared, so the innitial weight would be zero. Now, let's answer this by parts.

a) mass of displaced water.

In this case all we need to do is to substract the 0.70 with the 0.13 g. so:

mW = 0.70 - 0.13

mW = 0.57 g of water

b) Volume of water.

In this case, we have the density of water, so we use the formula for density and solve for volume:

d = m/V

V = m/d

Replacing:

Vw = 0.57/0.9982

Vw = 0.5710 mL of water

c) volume of the metal weight

In this case the volume would be the volume displaced of water, which would be 0.5710 mL

d) the mass of the metal weight.

In this case, it would be the mass when the metal weight hits the bottom which is 0.70 g

e) density.

using the above formula of density we calculate the density of the metal

d = 0.70 / 0.5710

d = 1.2259 g/mL

4 0
2 years ago
During this reaction, p4 + 5o2 → p4o10, 0.800 moles of product was made in 15.0 seconds. what is the rate of reaction? 56.8 g/mi
Eduardwww [97]
The solution for this problem would be:
The mass of P4O10 is computed by: 0.800 mol x 284 g/mol = 227g t = 15.0 s ( 1 min / 60 s) = 0.25 min 
So solving for the rate will be mass over t = m/t = 227/0.25 = 908 g/min would be the answer for this problem.
3 0
1 year ago
"A sample of silicon has an average atomic mass of 28.084amu. In the sample, there are three isotopic forms of silicon. About 92
adoni [48]

<u>Answer:</u> The percentage abundance for _{14}^{30}\textrm{Si} isotope is 3.09 %.

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i    .....(1)

We are given:

Let the fractional abundance for _{14}^{28}\textrm{Si} isotope be 'x'

  • <u>For _{14}^{28}\textrm{Si} isotope:</u>

Mass of _{14}^{28}\textrm{Si} isotope = 27.9769 amu

Percentage abundance of _{14}^{28}\textrm{Si} isotope = 92.22 %

Fractional abundance of _{14}^{28}\textrm{Si} isotope = 0.9222

  • <u>For _{14}^{29}\textrm{Si} isotope:</u>

Mass of _{14}^{28}\textrm{Si} isotope = 28.9764 amu

Percentage abundance of _{14}^{28}\textrm{Si} isotope = 4.68%

Fractional abundance of _{14}^{28}\textrm{Si} isotope = 0.0468

  • <u>For _{14}^{30}\textrm{Si} isotope:</u>

Mass of _{14}^{30}\textrm{Si} isotope = 29.9737 amu

Fractional abundance of _{14}^{30}\textrm{Si} isotope = x

  • Average atomic mass of silicon = 28.084 amu

Putting values in equation 1, we get:

28.084=[(27.9769\times 0.9222)+(28.9764\times 0.0468)+(29.9737\times x)]\\\\x=0.0309

Converting this fractional abundance into percentage abundance by multiplying it by 100, we get:

\Rightarrow 0.0309\times 100=3.09\%

Hence, the percentage abundance for _{14}^{30}\textrm{Si} isotope is 3.09 %.

6 0
2 years ago
Describe the process used to measure out a specific mass of a solid
Svetlanka [38]
Although the process varies slightly from one material to another, the general process is as follows:

1) Choose an appropriate container for the solid. This may be a petri dish or a beaker in which you want to prepare the solution of the solid or any other lab equipment.

2) Place the container on a mass balance, then turn the balance on. The mass balance will automatically zero-out the mass of the container, so that any mass that you add on the container will be the mass of the solid. Alternatively, you may first measure the mass of the empty container alone.

3) Add the solid using a lab spatula. The solid should be added more slowly when the reading on the scale comes close to the desired value.

4) Remove the container from the mass balance after the desired amount of solid has been added.
7 0
1 year ago
Other questions:
  • What is the percent composition of a solution in which 480 grams of sodium chloride, NaCl, is dissolved in 4 liters of solution.
    12·1 answer
  • Which has more momentum- an 80 kg runner with a velocity of 2.45 m/s or a 65 kg runner with velocity of 3 m/s? Show work
    8·1 answer
  • How many atoms of oxygen are contained in 160 grams of N2O3?
    13·2 answers
  • Carbon dioxide readily absorbs radiation with an energy of 4.67 x 10-20 J. What is the wavelength and frequency of this radiatio
    10·1 answer
  • NH4NO3, whose heat of solution is 25.7 kJ/mol, is one substance that can be used in cold pack. If the goal is to decrease the te
    15·1 answer
  • How many grams o an ointment base must be added to 45 g o clobetasol (T EMOVAT E) ointment, 0.05% w/w, to change its strength to
    9·1 answer
  • Now imagine you have several of such dipoles, and place them regularly between the plates. For this part of the pre-lab, you can
    8·1 answer
  • The successive ionization energies of a certain element are I1 = 589.5 kJ/mol, I2 =1145 kJ/mol, I3= 4900 kJ/mol, I4 = 6500 kJ/mo
    8·2 answers
  • Dr. Franck cuts a bar of pure gold into smaller and smaller pieces. Will this action change the element that makes up the bar? E
    10·1 answer
  • In an NMR experiment, shielding refers to the reduced impact of the ______on a nucleus due to the presence of ______around the n
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!