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Olegator [25]
3 years ago
6

PLEASE HELP ASAP!!!! WILL GIVE BRANILEST

Chemistry
1 answer:
xenn [34]3 years ago
4 0

Answer:

100 ml 108 g would be the answer.

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What passes through the Haversian canal
Tamiku [17]

A Haversian canal is referred to as any minute tube, which forms a network in a bone and also contains blood vessels. Blood vessels, nerve fibers and lymphatics may pass through the Haversian canal. Haversian canals are found surrounding blood vessels and nerve cells all over the bones and interacts with bone cells from end to end networks known as canaliculi.

4 0
3 years ago
. Calculate the final Celsius temperature of sulfur dioxide gas if 50.0 mL of the gas at 20 C and 0.450 atm is heated until the
Anna [14]

Answer:

The final temperature of sulfur dioxide gas is 215.43 C

Explanation:

Gay Lussac's Law establishes the relationship between the temperature and the pressure of a gas when the volume is constant. This law says that if the temperature increases the pressure increases, while if the temperature decreases the pressure decreases. In other words, the pressure and temperature are directly proportional quantities.

Mathematically, the Gay-Lussac law states that, when a gas undergoes a transformation at constant volume, the quotient of the pressure exerted by the temperature of the gas remains constant:

\frac{P}{T}=k

Assuming you have a gas that is at a pressure P1 and at a temperature T1 at the beginning of the experiment, by varying the temperature to a new value T2, then the pressure will change to P2, and it will be true:

\frac{P1}{T1} =\frac{P2}{T2}

The reference temperature is the absolute temperature (in degrees Kelvin)

In this case:

  • P1= 0.450 atm
  • T1= 20 C= 293.15 K (being 0 C= 273.15 K)
  • P2=0.750 atm
  • T2= ?

Replacing:

\frac{0.450atm}{293.15 K} =\frac{0.750 atm}{T2}

Solving:

T2 =\frac{0.750 atm}{\frac{0.450atm}{293.15 K} }

T2=\frac{0.750 atm}{0.450 atm} *293.15K

T2=488.58 K

Being 273.15 K= 0 C, then 488.58 K= 215.43 C

<u><em>The final temperature of sulfur dioxide gas is 215.43 C</em></u>

6 0
3 years ago
How can you tell from the name the types of bonds present in a hydrocarbon?
Aneli [31]

Answer:

Of bond is formed by sharing of electrons - covalent

If bond is formed by donation and by accepting electrons - ionic bond

Bond formed between metals - metallic bond.

7 0
3 years ago
Read 2 more answers
Write a balanced chemical equation for the reaction that occurs when heptane, C7H16(l) , burns in air.
Pavlova-9 [17]
Start with Unbalanced Equation and balance it, so...
C7H16+O2--->CO2+H2O
There are 7 C atoms on the left-hand side, so we need 7 C atoms on the right-hand side. Add a 7 in front of the CO2...7CO2+H2O on right side now.
We have fixed 16 H atoms on the left-hand side, so we need 16 H atoms on the right-hand side. Add an 8 in front of H2O to make 16 (8x2)...7CO2+8H2O on right side now.
There are 22 O atoms on the right-hand side: 14 from the CO2 and 8 from the H2O. Add an 11 in front of the O2 on the left side to make 22 (11x2).
Every formula now has a fixed coefficient. You should have a balanced equation of...
C7H16+11O2--->7CO2+8H2O
8 0
3 years ago
Would you expect the wavelength of maximum absorbance for [Cu(NH3)4]2+ to be greater than or less than the wavelength of maximum
DENIUS [597]

Answer:

Less

Explanation:

Since [Cu(NH3)4]2+ and [Cu(H2O)6]2+ are Octahedral Complexes the transitions between d-levels explain the majority of the absorbances seen in those chemical compounds. The difference in energy between d-levels is known as ΔOh (ligand-field splitting parameter) and it depends on several factors:

  • The nature of the ligand: A spectrochemical series is a list of ligands ordered on ligand strength. With a higher strength the ΔOh will be higher and thus it requires a higher energy light to make the transition.
  • The oxidation state of the metal: Higher oxidation states will strength the ΔOh because of the higher electrostatic attraction between the metal and the ligand

A partial spectrochemical series listing of ligands from small Δ to large Δ:

I− < Br− < S2− < Cl− < N3− < F−< NCO− < OH− < C2O42− < H2O < CH3CN < NH3 < NO2− < PPh3 < CN− < CO

Then NH3 makes the ΔOh higher and it requires a higher energy light to make the transition, which means a shorter wavelength.

7 0
3 years ago
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