The answer to your question is A. Wrought iron because amongst its other properties, wrought iron becomes soft at red heat, and can be easily be forged and forge welded.
Answer:
Option-B (Carbon and Silicon)
Explanation:
Among the given pairs only carbon and silicon have the most similar properties. This is because,
Sodium and Magnesium belong to different groups. Sodium present in Group I has one electron in its valence shell and capable of transferring only one electron while, Magnesium present in Group II have two electrons in its valence shell and is capable of donating two electrons. Hence, both show different properties.
Example:
2 Na + Cl₂ → NaCl
Mg + Cl₂ → MgCl₂
As shown in reactions when Sodium and Magnesium are treated with Cl₂ they give a products with different proportions.
Carbon and Silicon show almost same properties because both belong to Group IV hence both are capable of forming four bonds. Also, they share the same property of self linkage in making a long chains.
Argon and Chlorine also belong to two different groups. Argon is present in Group VIII (Noble Gases) and Chlorine is present in Group VII (Halogens). Hence, Argon is an inert specie which is non reactive while Chlorine gives different reaction easily.
Potassium and Calcium belong to different groups. Potassium present in Group I has one electron in its valence shell and capable of transferring only one electron while, Calcium present in Group II have two electrons in its valence shell and is capable of donating two electrons. Hence, both show different properties.
Example:
2 K + Cl₂ → KCl
Ca + Cl₂ → CaCl₂
As shown in reactions when Potassium and Calcium are treated with Cl₂ they give a products with different proportions.
Answer:
Cyclopropane has a planar carbon back bone while propane does not
Explanation:
We have to recognize that in straight chain saturated organic compounds, carbon atoms have a tetrahedral geometry. Each carbon atom is bonded to four other atoms.
However, carbon atoms in cyclic compounds are also sp3 hybridized with each carbon bonded to only four other atoms but the ring system is highly strained.
Cyclopropane is a necessarily planar molecule with a bond angle that is far less than the expected tetrahedral bond angle due to strain in the molecule. Hence, the carbon atoms may have have a "planar backbone".
Answer:
B) THE DEPTH OF THE LAKE IS 0.060 m
Explanation:
b) Determine the depth of the lake in metres
1. Using the general gas law, we will calculate the initial pressure of the air bubbles.
P1V1 /T1 = P2V2/T2
P1 = Unknown
T1 = 5.24 °C
T2 = 18.73 °C
P2 = 0.973 atm
V1 = V1
V2 = 6V1
P1 = P2 V2 T1 / V1 T2
P1 = 0.973 * 6V1 * 5.24 / V1 * 18.73
P1 = 5.09852 * 6 / 18.73
P1 = 30.59112 / 18.73
P1 = 1.633 atm.
2. Calculate the depth of the lake:
Pressure = length * density * acceleration
length = Pressure / density * acceleration
Pressure = 1.633 atm = 1.633 * 101, 325 Nm^2 = 165, 463.725 Nm^2
Density = 1.02 g/cm3 = 1.02 * 10^3 kg/m^3
Acceleration = 9.8 m/s^2
So therefore, the length in metres is:
Length = density * acceleration / pressure
Length = 1.02 *10^3 * 9.8 / 165, 463.725
Length = 9.996 * 10^3 / 165 463.725
Length = 0.06 m
Hence, the depth of the lake is 0.06 m
