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Sedaia [141]
3 years ago
14

Arrange the events in the life of a star in the correct order.

Chemistry
1 answer:
tatiyna3 years ago
3 0

Answer:

1. The temperature increases in the center of a nebula, and nuclei begin to fuse.

2. As the nuclear fusion sets off a chain reaction, the nebula gets hotter and brighter and turns into a star.

3. The star stays in the main sequence stage for a long time.

4. The star runs out of hydrogen and turns into a planetary nebula or a supernova, depending on its mass.

5. Heavy elements are created when the supernova explodes.

6. All that remains of the star now is the core. In some cases, the star turns into a black hole.

Explanation:

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How many moles of sodium carbonate are contained by 57.3g of sodium carbonate
Lady_Fox [76]

Answer:

\boxed {\boxed {\sf 0.541 \  mol \ Na_2CO_3}}

Explanation:

We are asked to find how many moles of sodium carbonate are in 57.3 grams of the substance.

Carbonate is CO₃ and has an oxidation number of -2. Sodium is Na and has an oxidation number of +1. There must be 2 moles of sodium so the charge of the sodium balances the charge of the carbonate. The formula is Na₂CO₃.

We will convert grams to moles using the molar mass or the mass of 1 mole of a substance. They are found on the Periodic Table as the atomic masses, but the units are grams per mole instead of atomic mass units. Look up the molar masses of the individual elements.

  • Na:  22.9897693 g/mol
  • C: 12.011 g/mol
  • O: 15.999 g/mol

Remember the formula contains subscripts. There are multiple moles of some elements in 1 mole of the compound. We multiply the element's molar mass by the subscript after it, then add everything together.

  • Na₂ = 22.9897693 * 2= 45.9795386 g/mol
  • O₃ = 15.999 * 3= 47.997 g/mol
  • Na₂CO₃= 45.9795386 + 12.011 + 47.997 =105.9875386 g/mol

We will convert using dimensional analysis. Set up a ratio using the molar mass.

\frac {105.9875386  \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}

We are converting 57.3 grams to moles, so we multiply by this value.

57.3 \ g \ Na_2CO_3} *\frac {105.9875386  \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}

Flip the ratio so the units of grams of sodium carbonate cancel.

57.3 \ g \ Na_2CO_3} *\frac {1 \ mol \ Na_2CO_3}{105.9875386  \ g \ Na_2CO_3}

57.3 } *\frac {1 \ mol \ Na_2CO_3}{105.9875386 }

\frac {57.3 }{105.9875386 } \ mol \ Na_2CO_3

0.5406295944 \ mol \ Na_2CO_3

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we found that is the thousandth place. The 6 in the ten-thousandth place to the right tells us to round the 0 up to a 1.

0.541 \  mol \ Na_2CO_3

There are approximately <u>0.541 moles of sodium carbonate</u> in 57.3 grams.

6 0
3 years ago
The ph of a 0.55 m aqueous solution of hypobromous acid, hbro, at 25.0 °c is 4.48. what is the value of ka for hbro?
gladu [14]
Hbro  dissociate  as  follows
HBro--->  H+  +  BrO-
  Ka=  (H+)(BrO-)  /  HBro
 PH  =  -log (H+)
therefore (H+)  =  10^-4.48= 3.31  x10^-5
ka is  therefore= ( 3.31  x 10^-5)^2/0.55=1.99  x10^-9
6 0
3 years ago
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