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yan [13]
2 years ago
13

The van der waals equation is a modified version of the

Chemistry
1 answer:
vodomira [7]2 years ago
6 0

Answer:

Hullbreine

Explanation:

answer from quiz

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Neve later believes that New Zealand's winter is caused by the sun circling to the other side
VARVARA [1.3K]

Explanation:

I need a help of math.

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2 years ago
In a pure metal, the electrons can be thought of as [ Select ] throughout the metal. Using molecular orbital theory, there [ Sel
Phoenix [80]

Answer:

Explanation:

In a pure metal, the electrons can be thought of as [concentrated] around atoms throughout the metal. Using molecular orbital theory, there [is ] an energy gap between the filled molecular orbitals and empty molecular orbitals. The [antibonding] orbitals are typically higher in energy and are mostly (filled]

5 0
2 years ago
Use the drop-down menus to select the correct name for each of the organic compounds.
Molodets [167]

Answer:

Here's what I get  

Explanation:

CH₃CH₂CH₂CH₂CH₂CH₃ —  hexane

CH₂=CHCH₂CH₂CH₂CH₃ —  hex-1-ene is the preferred IUPAC name (PIN). 1-Hexene is accepted

CH₃C≡CCH₃ —  but-2-yne (PIN); 2-butyne is accepted

CH₃CH(CH₃)CH₂CH₂CH₃ — 2-methylpentane

CH₃CH₂CHCICH₂CH₃ — 3-chloropentane

5 0
3 years ago
Read 2 more answers
The electronic configuration of phosphorous is 2.8.5. Explain, in terms of its electronic configuration, why phosphorus is in gr
Setler [38]

Answer:

Explanation:

phosphorus belongs to group 5 of the periodic table because it has 5 electron in its outermost shell the number of electron in the outermost shell of electron determine the group of the element in the periodic table

6 0
3 years ago
Copper has a specific heat of 0.385 J/gºC.
Anna71 [15]

Answer:

The final temperature is 348.024°C.

Explanation:

Given data:

Specific heat of copper = 0.385 j/g.°C

Energy absorbed = 7.67 Kj (7.67×1000 = 7670 j)

Mass of copper = 62.0 g

Initial temperature T1 = 26.7°C

Final temperature T2 = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

Q = m.c. ΔT

7670 J = 62.0 g × 0.385  j/g °C ×( T2- 26.7 °C )

7670 J = 23.87 j.°C ×( T2- 26.7 °C )

7670 J / 23.87 j/°C = T2- 26.7 °C

T2- 26.7 °C = 321.324°C

T2 = 321.324°C + 26.7 °C

T2 = 348.024°C

The final temperature is 348.024°C.

6 0
3 years ago
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