Answer:
2H2O2-----2H2O+O2
Explanation:
This is because theres the same number of atoms of each element on both sides
Is this the whole answer?
Answer:
1.7x10^8 Hz
Explanation:
Frequency could be explained as the number of occurrences of a repeating event at a time
Given:
wavelength = 1.8 meters
The frequency f of the waves can be calculated using f = c / λ
Where c (m/s) is the speed of the wave
λ (m) is the wavelength
Speed c= 3*10^8 m/s
Frequency f= 3*10^8 /1.8
Frequency= 1.7x10^8 Hz
Therefore,the frequency of waves from a radar detector is 1.7x10^8 Hz
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.
Combustion equation of n-hexane:
2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O
a)
Assuming we have 100 moles of air,
Oxygen = 20.9 moles
n-hexane required = 20.9/19 x 2
= 2.2 moles
LFL = Half of stoichometric amount = 2.2 / 2 = 1.1
LFL n-hexane = 1.1%
b)
1.1 volume percent required for LFL
1.1% x 1
= 0.0011 m³ of n-hexane required
