Its phosphorus (P)In writing the electron configuration for Phosphorus the first two electrons will go in the 1s orbital. Since 1s can only hold two electrons the next 2 electrons for Phosphorous go in the 2s orbital. The next six electrons will go in the 2p orbital. The p orbital can hold up to six electrons. We'll put six in the 2p orbital and then put the next two electrons in the 3s. Since the 3s if now full we'll move to the 3p where we'll place the remaining three electrons. Therefore the Phosphorus electron configuration will be 1s22s22p63s23p3.
Properties of a solution that depend only on the ratio of the number of particles of solute and solvent in the solution are known as colligative properties. For this problem, we use boiling point elevation concept.
ΔT(boiling point) = (Kb)mi
ΔT(boiling point) = (0.51 C-kg / mol )(4.0 mol / 2.05 kg ) (2)
ΔT(boiling point) = 1.99 C
T(boiling point) = 101.99 C
Answer:
The volume will decrease and the balloons will be smaller
Answer:
0.296 J/g°C
Explanation:
Step 1:
Data obtained from the question.
Mass (M) =35g
Heat Absorbed (Q) = 1606 J
Initial temperature (T1) = 10°C
Final temperature (T2) = 165°C
Change in temperature (ΔT) = T2 – T1 = 165°C – 10°C = 155°C
Specific heat capacity (C) =..?
Step 2:
Determination of the specific heat capacity of iron.
Q = MCΔT
C = Q/MΔT
C = 1606 / (35 x 155)
C = 0.296 J/g°C
Therefore, the specific heat capacity of iron is 0.296 J/g°C
