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Alenkasestr [34]
2 years ago
11

Is 6.0 × 10 to the negative 4th power is a significant figure?​

Chemistry
1 answer:
Dmitry_Shevchenko [17]2 years ago
6 0

Answer:

Yes

Explanation:

First remember that a significant figure are basically values that contribute to the precision of a value. In any scientific notation the values are significant figures because these values stay the same. In this case we have two significant figures which is 6 and 0, you can further prove that they are significant figures by converting the notation into standard form.

6.0\times10^{-4}

Negative so move the decimal point to the left:

=0.00060

6 and 0 are the significant figures in this standard notation because it's precise to it's actual value which is 6.0.

To sum up, the values you have on the left side of a notation are significant figures since they will not change no matter if it's standard or scientific notation meaning it's precise.

Hope this helps.

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Question 9 (1 point)
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Noble gases

Explanation:

     Electronic configuration 1s²  2s²   2p⁶

 The element belongs to the group of the noble gases.

  • The noble gases have complete outer shell configuration of their atoms.
  • we can infer that the configuration above is for an element in the p-block because the last sub-level filled is the p-orbital.
  • The elements therefore belongs to the p-block
  • The block is from group 111A to O
  • Only the halogens and noble gases fits this picture from the option.
  • The outer most p-subshell have three orbitals requiring 6 electrons to fill them up.
  • This makes a complete and stable configuration.
  • The highest energy level of 2 is also made up of 8 electrons, an octet.
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AnnZ [28]

Answer:

Oxide sugar dissolved in water pond

because water + sugar= sugar melts but the oxide purifys the water.

Explanation:

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6 0
2 years ago
12. How are secondary sources of energy different from other energy resources?<br> (Plz help)
Pani-rosa [81]

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(04.01 LC)
Marysya12 [62]

Answer:

Substance at the beginning of a reaction- reactant

Substance at the end of a reaction- product

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Explanation:

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