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Daniel [21]
4 years ago
14

Give the number of significant figures: 0.01101 cm Significant figures:

Chemistry
1 answer:
pantera1 [17]4 years ago
7 0

The answer you are looking for is 4.


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How many formula units of CaO are in 32.7 g of CaO?
zhuklara [117]

Answer:

D) 3.51x10²³ formula units

Explanation:

To know this, we need to calculate the molecular mass of CaO. To do that, let's use the atomic mass of each element of this compound.

Ca: 40.08 g/mol;    O: 16 g/mol

With these given AM, let's calculate the molar mass of CaO:

Mm CaO = 40.08 + 16 = 56.08 g/mol

Now that we have the MM of CaO, it's time to determine the moles of CaO:

moles = mass / MM

moles = 32.7 / 56.08 = 0.5831 moles

Finally to get the formula units, we need to use the Avogadro's number which is 6.02x10²³. According to this number, 1 mole of any substance has this number of atoms, molecules or formula units. So, if we have 0.5831 moles, then the formula units will be:

FU = 0.5831 * 6.02x10²³

<h2>FU = 3.51x10²³ formula units</h2>

Hope this helps

7 0
3 years ago
Please help! Thanks!
olya-2409 [2.1K]

Answer:

<u>The deviations are :</u>

  • <u>The activation energy which changes with temperature</u>
  • <u>The arrhenius constant which depends on the temperature</u>

Explanation:

  • There are deviations from the Arrhenius law during the glass transition in all classes of glass-forming matter.
  • The Arrhenius law predicts that the motion of the structural units (atoms, molecules, ions, etc.) should slow down at a slower rate through the glass transition than is experimentally observed.
  • In other words, the structural units slow down at a faster rate than is predicted by the Arrhenius law.
  • <em>This observation is made reasonable assuming that the units must overcome an energy barrier by means of a thermal activation energy. </em>
  • The thermal energy must be high enough to allow for translational motion of the units <em>which leads to viscous flow of the material.</em>

  • Both the Arrhenius activation energy and the rate constant k are experimentally determined, and represent macroscopic reaction-specific parameters <em>that are not simply related to threshold energies and the success of individual collisions at the molecular level. </em>
  • Consider a particular collision (an elementary reaction) between molecules A and B. The collision angle, the relative translational energy, the internal (particularly vibrational) energy will all determine the chance that the collision will produce a product molecule AB.
  • Macroscopic measurements of E(activation energy) and k(rate constant ) <em>are the result of many individual collisions with differing collision parameters. </em><em>They are averaged out to a macroscopic quantity.</em>
6 0
3 years ago
I need help with my lab report
stich3 [128]

Answer:

learn more. feather learning can help

7 0
3 years ago
(06.01 LC)
Brums [2.3K]

Answer:

Sorry; I am unfamiliar with the details they taught you. I would guess #3, for it allows kinetics to “work on the particles”. #1 makes no sense; what is

“Large”? #2 infers all particles are charged....#4. Seems unclear to me.

Maybe these thoughts will jar your memory.

Explanation:

4 0
4 years ago
Why is DNA an important molecule in the cell?
vagabundo [1.1K]
DNA holds all the genetic information for a cell. It helps with basically all cell processes.
5 0
4 years ago
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