B
Explanation:
Burning is a chemical change and cutting grass is a physical change
Answer:
Noble gas Electronic configuration of arsenic:
As₃₃ = [Ar] 3d¹⁰ 4s² 4p³
Explanation:
Arsenic is metalloid.
Its atomic number is 33.
Its atomic mass is 75 amu.
Its symbol is As.
It is usually present in combine with sulfur and metals.
it is used in bronzing.
It is also used for hardening.
Electronic configuration:
As₃₃ = Is² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p³
Noble gas Electronic configuration:
As₃₃ = [Ar] 3d¹⁰ 4s² 4p³
Noble gas electronic configuration is shortest electronic configuration by using the noble gas elements full octet electronic configuration.
Answer:
1. V2.
2. 299K.
3. 451K
4. 0.25 x 451 = V2 x 299
Explanation:
1. The data obtained from the question include:
Initial volume (V1) = 0.25mL
Initial temperature (T1) = 26°C
Final temperature (T2) = 178°C
Final volume (V2) =.?
2. Conversion from celsius to Kelvin temperature.
T(K) = T (°C) + 273
Initial temperature (T1) = 26°C
Initial temperature (T1) = 26°C + 273 = 299K
3. Conversion from celsius to Kelvin temperature.
T(K) = T (°C) + 273
Final temperature (T2) = 178°C
Final temperature (T1) = 178°C + 273 = 451K
4. Initial volume (V1) = 0.25mL
Initial temperature (T1) = 299K
Final temperature (T2) = 451K
Final volume (V2) =.?
V1 x T2 = V2 x T1
0.25 x 451 = V2 x 299
Explanation:
The valence electrons within an atom is the number of electrons in its outermost shell.
These electrons are used by an atom to react with one another. They determine the extent to which an atom is ready to combine either by losing, gaining or sharing these electrons.
- Every atom desires to have a completely filled outermost shell.
- Only the elements in group 8 have a complete octet.
- The need to attain stability is driven by the number of electrons in their valence shell.
- Therefore, some atoms are very reactive.
- Those needing one electrons to complete their octet and also those that must lose one electron are very reactive.
Answer:
mass P4 = 35.998 g
Explanation:
∴ STP: P = 1 atm; T = 298 K
∴ V O2= 35.5 L
⇒ nO2 = P.V / R.T
∴ R = 0.082 atm.L/K.mol
⇒ nO2 = ((1 atm)×(35.5L))/((0.082 atm.L/K.mol)(298K))
⇒ nO2 = 1.453 mol O2
⇒ mol P4 = (1.453 molO2)×(mol P4/ 5molO2) = 0.2906 mol P4
∴ Mw P4 = 123.895 g/mol
⇒ mass P4 = (0.2906 mol P4)×(123.895 g/mol) = 35.998 g P4
