Answer:
When the volume will be reduced to 2.50 L, the temperature will be reduced to a temperature of 230.9K
Explanation:
Step 1: Data given
A sample of sulfur hexafluoride gas occupies a volume of 5.10 L
Temperature = 198 °C = 471 K
The volume will be reduced to 2.50 L
Step 2 Calculate the new temperature via Charles' law
V1/T2 = V2/T2
⇒with V1 = the initial volume of sulfur hexafluoride gas = 5.10 L
⇒with T1 = the initial temperature of sulfur hexafluoride gas = 471 K
⇒with V2 = the reduced volume of the gas = 2.50 L
⇒with T2 = the new temperature = TO BE DETERMINED
5.10 L / 471 K = 2.50 L / T2
T2 = 2.50 L / (5.10 L / 471 K)
T2 = 230.9 K = -42.1
When the volume will be reduced to 2.50 L, the temperature will be reduced to a temperature of 230.9K
Answer: 3.75 M
Explanation:
400 mL = 0.4 L
NaOH has a molar mass of around 40 g/mol.
= 1.5 moles
Molarity = 
= 3.75 M
Answer:
d. Radon-222
Explanation:
²²⁶₈₈Ra → ²²²₈₆Rn + ⁴₂He
Alpha particle is a helium nucleus with mass number 4 and atomic number 2. According to the law of conversation of mass, the sum of the mass number and atomic number must be equal on both side of the reaction.
Since the mass number of Ra is 226 and that of He is 4. The mass number of the unknown element must be 226 - 4 = 222.
Since the atomic number of Ra is 88 and that of He is 2. The atomic number of the unknown element must be 88 - 2 = 86.
Now looking in the periodic table Radon is the only element with atomic number 86.
For the given molecule, we are asked to give-
- The electron configuration of an isolated B atom
- The electron configuration of an isolated F atom
- Hybrid orbitals should be constructed on the B atom to make the B–F bonds in Boron tri flouride
- valence orbitals, if any, remain unhybridized on the B atom.
- The electron configuration of an isolated B atom:
as atomic number of B is 5
electronic configuration will be [He] 2s² 2p¹
- The electron configuration of an isolated F atom:
as atomic number of F is 9
electronic configuration will be [He] 2s² 2p5
- Hybrid orbitals should be constructed on the B atom to make the B–F bonds in Boron tri flouride will be sp2.
as the one s and two of p orbital from the valance shell will hybridised to make 3 hybrid orbital of B resulting in 3 B-F bonds.
- valence orbitals, if any, remain unhybridized on the B atom will be 1
To know more about hybrisisation:
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Answer:
increase , my age increase
Explanation:
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