<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.
Answer:
.025 ml 02
Explanation:
I made a pdf and pasted it in and etc
6CO2 + 6H2O → C6H12O6 + 6O2
glucose produced = 1/6 x 23.6 = 3.93 moles
Answer:
Explanation:
Hygrometer, instrument used in meteorological science to measure the humidity, or amount of water vapour in the air. Several major types of hygrometers are used to measure humidity.
Answer:
7.04 g
Explanation:
Let's consider the reaction in the last step of the Ostwald process.
3 NO₂(g) + H₂O(l) → 2 HNO₃(aq) + NO(g)
The molar mass of HNO₃ is 63.01 g/mol. The moles corresponding to 6.40 g are:
6.40 g × (1 mol/63.01 g) = 0.102 mol
The molar ratio of NO₂ to HNO₃ is 3:2. The reacting moles of NO₂ are:
0.102 mol HNO₃ × (3 mol NO₂/2 mol HNO₃) = 0.153 mol NO₂
The molar mass of NO₂ is 46.01 g/mol. The mass corresponding to 0.153 moles is:
0.153 mol × (46.01 g/mol) = 7.04 g
