Answer:
The answer is:
B
Explanation:
The compound in Option B is Methane.
Methane is known to be a compound which has two elements, carbon and hydrogen. It has a central atom which is surrounded by four hydrogen atoms. It's chemical formula is CH4.
Methane's outer atoms are dipoles and are in the same direction. This makes the overall molecule non-polar. The compound itself has non-polar bonds and it is non-polar itself.
The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
Average bond enthalpy = 594.9kJ/mol/2
=297.45kJ/mol
∴ Averange bond enthalpy of O3(g) is 297.45kJ/mol.
Answer:
Explanation:
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Answer:
There are 1.05 x 10²⁴ molecules in 48.6 g N₂
Explanation:
1 mol of N₂ has a mass of (14 g * 2) 28 g.
Then, 48.6 g of N₂ will be equal to (48.6 g *(1 mol/ 28 g)) 1.74 mol.
Since there are 6.022 x 10²³ molecules in 1 mol N₂, there will be
(1.74 mol *( 6.022 x 10²³ / 1 mol)) 1.05 x 10²⁴ molecules in 1.74 mol N₂ (or 48. 6 g N₂).
<h3>
Answer:</h3>
83.33 seconds.
<h3>
Explanation:</h3>
<u>We are given;</u>
- Take off velocity as 300 km/hr
- Acceleration as 1 m/s²
We are required to calculate the take off time of the airplane.
<h3>Step 1: Convert velocity from km/hr to m/s </h3>
We are going to use the conversion factor.
The conversion factor is 3.6 km/hr per m/s
Therefore;
Velocity = 300 km/hr ÷ 3.6 km/hr per m/s
= 83.33 m/s
<h3>Step 2: Calculate the take off time</h3>
We know that;
v = u + at
where, u is the initial velocity, v the final velocity, a the acceleration and t is time.
But, initial velocity is Zero
Therefore;
83.33 m/s = 1 m/s² × t
Thus;
time = 83.33 m/s ÷ 1 m/s²
= 83.33 seconds
Therefore, the take off time is 83.33 seconds.
