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Jobisdone [24]
3 years ago
11

The molar mass of aluminum oxide (Al2O3) is 102.0 g/mol. What is the correct way to write the inverted molar mass of aluminum ox

ide as a conversion factor?
Chemistry
2 answers:
rosijanka [135]3 years ago
8 0

Answer:

B on edge

Explanation:

ki77a [65]3 years ago
6 0
The molar mass of aluminum oxide is 102.0 g/mol =>

     1 mol Al2 O3 = 102.0 g

=> 1 = 1 mol Al2O3 / 102.0 g = 1 / 102.0 g/mol

So, the conversion factor, which you can use to convert mass in grams of aluminum oxide to number of moles is 1 / 102.0 g/mol.

Answer: 1 / 102.0 g/mol

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A 1. 07 g sample of a noble gas occupies a volume of 363 ml at 35°c and 678 mmhg. Identify the noble gas in this sample. (r = 0.
Margaret [11]

The identity of the noble gas is the sample is Krypton

<h3>Ideal Gas law</h3>

From the question, we are to determine the identity of the noble gas in the sample

From the ideal gas equation, we have that

PV = nRT

∴ n = PV / RT

Where P is the pressure

V is the volume

n is the number of moles

R is the gas constant

and T is the temperature

From the given information,

P = 678 mmHg = 0.892105 atm

V = 363 mL = 0.363 L

R = 0.08206 L.atm/mol.K

T = 35 °C = 35 + 273.15 K = 308.15 K

Putting the parameters into the equation, we get

n = (0.892105 × 0.363)/ (0.08206 × 308.15)

n = 0.0128 moles

Now, we will determine the Atomic mass of the sample

Using the formula,

Atomic = Mass / Number of moles

Atomic mass of the substance = 1.07 / 0.0128

Atomic mass of the substance = 83.6 amu

The noble gas with the closest atomic mass to this value is Krypton.

Molar mass of Krypton = 83.798 amu

Hence, the identity of the noble gas is the sample is Krypton

Learn more on Ideal Gas law here: brainly.com/question/20212888

#SPJ12

4 0
2 years ago
What is the mole ratio of the following equation?
Liono4ka [1.6K]
The mole ratio would be 2:2:2
7 0
3 years ago
How can you increase friction
Lelu [443]

Answer:Well There is many ways to increase friction between objects. One being rubbing the two objects together quicker and harder. If there is any sort of wetness or anything related to that make sure to dry the surface between the two objects you want to create friction between so it will be more effective.

Explanation:

5 0
3 years ago
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How many milliliters of a 3.4 M NaCl solution would be needed to prepare each solution?
Ksenya-84 [330]

Answer:

a. Approximately 1.3\; \rm mL.

b. Approximately 7.2\; \rm mL.

Explanation:

The unit of concentration "\rm M" is equivalent to "\rm mol \cdot L^{-1}", which means "moles per liter."

However, the volume of both solutions were given in mililiters \rm mL. Convert these volumes to liters:

\displaystyle 45\; \rm mL = 45\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.045\; \rm L.

\displaystyle 330\; \rm mL = 330\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.330\; \rm L.

In a solution of volume V where the concentration of a solute is c, there would be c \cdot V (moles of) formula units of this solute.

Calculate the number of moles of \rm NaCl formula units in each of the two solutions:

Solution in a.:

n = c \cdot V = 0.045\; \rm L \times 0.10\; \rm mol \cdot L^{-1} = 0.0045\; \rm mol.

Solution in b.:

n = c \cdot V = 0.330\; \rm L \times 0.074\; \rm mol \cdot L^{-1} = 0.02442\; \rm mol.

What volume of that 3.4\; \rm M (same as 3.4 \; \rm mol \cdot L^{-1}) \rm NaCl solution would contain that many

For the solution in a.:

\displaystyle V = \frac{n}{c} = \frac{0.0045\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0013\; \rm L.

Convert the unit of that volume to milliliters:

\displaystyle 0.0013\; \rm L = 0.0013\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 1.3\; \rm mL.

Similarly, for the solution in b.:

\displaystyle V = \frac{n}{c} = \frac{0.02442\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0072\; \rm L.

Convert the unit of that volume to milliliters:

\displaystyle 0.0072\; \rm L = 0.0072\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 7.2\; \rm mL.

8 0
3 years ago
What happens when an electron jumps energy from energy level 3 energy level 6 in an atom
Alik [6]

Explanation:

high energy to Low energy

=the electron gains energy (K.E)

4 0
2 years ago
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