Fill in blanks plz help 23 points

Chemistry

1 answer:

shtirl [24]2 years ago

6 0

I'm not sure about the second one so sorry for that! But, the other three blanks are length, graduated cylinder, and matter (I believe its matter, I'm sorry if thats wrong)

Again, I'm sorry I couldn't answer one and accurately answer the other one but I hope this helps you!

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Use the Periodic Table.

Zina [86]

Answer:

The answer to your question is: d

Explanation:

Electron configuration of Carbon.

Atomic number of carbon = 6

a) 1s²2s²2p¹ Number of electrons = 5, This is the electron configuration of Boron.

b) 1s²2s²2p⁴ Number of electrons = 8. This is the electron configuration of oxygen.

c) 1s²2s²p² Number of electrons = 6. This is the electron configuration of carbon. But I think is lacking a number two before p.

d) 1s²2s²2p² Number of electrons 6, This is the correct electron configuration of carbon.

4 0

2 years ago

A sample of neon occupies a volume of 752 ml at 25 0

lyudmila [28]

$V_{initial} = 752\:mL$
$T_{initial} = 25.0^0C$
converting to Kelvin
TK = TC + 273
TK = 25.0 + 273 → TK = 298.0 → $T_{initial} = 298.0\:K$
$V_{final} = ? (in\:milliliters)$
$T_{final} = 50.0^0C$
TK = TC + 273
TK = 50.0 + 273 → TK = 323.0 → $T_{final} = 323.0\:K$

By the first Law of Charles and Gay-Lussac, we have:
$\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }$

Solving:
$\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }$
$\frac{ 752 }{ 298.0 } = \frac{ V_{f} }{ 323.0 }$
Product of extremes equals product of means:
$298.0* V_{f} = 752*323.0$
$298.0 V_{f} = 242896$
$V_{f} = \frac{242896}{298.0}$
$\boxed{\boxed{V_{f} \approx 815.08\:mL}}\end{array}}\qquad\quad\checkmark$

5 0

3 years ago

Explain why molecules of water were attracted to the negatively charged plastic tube while molecules of cyclohexane were not. Dr

mihalych1998 [28]

Answer:

See attached

Explanation:

Structural geometry is the key to understanding particle-particle interactions on a physical levels. Suggest a review of molecular geometry and states of matter for a more in depth understanding. Doc