Answer:
THE FINAL TEMPREATURE OF IRON SWORD IS
1331°C
Answer:
Close to the calculated endpoint of a titration - <u>Partially open</u>
At the beginning of a titration - <u>Completely open</u>
Filling the buret with titrant - <u>Completely closed</u>
Conditioning the buret with the titrant - <u>Completely closed</u>
Explanation:
'Titration' is depicted as the process under which the concentration of some substances in a solution is determined by adding measured amounts of some other substance until a rection is displayed to be complete.
As per the question, the stopcock would remain completely open when the process of titration starts. After the buret is successfully placed, the titrant is carefully put through the buret in the stopcock which is entirely closed. Thereafter, when the titrant and the buret are conditioned, the stopcock must remain closed for correct results. Then, when the process is near the estimated end-point and the solution begins to turn its color, the stopcock would be slightly open before the reading of the endpoint for adding the drops of titrant for final observation.
Answer:
Both Options C and D are appropriate.
But I'd go with Option D since "Direct Air Capture" would eventually lead to "Ground Injection"
OPTION D.
Answer:
fundamental frequency in helium = 729.8 Hz
Explanation:
Fundamental frequency of an ope tube/pipe = v/2L
where v is velocity of sound in air = 340 m/s; λ is wave length of wave = 2L ; L is length of the pipe
To find the length of the pipe,
frequency = velocity of sound / 2L
272 = 340 / 2 L
L = 0.625 m
If the pipe is filled with helium at the same temperature, the velocity of sound will change as well as the frequency of note produced since velocity is directly proportional to frequency of sound.
Also, the velocity of sound is inversely proportional to square root of molar mass of gas; v ∝ 1/√m
v₁/v₂ = √m₂/m₁
v₁ = velocity of sound in air, v₂ = velocity of sound in helium, m₁ = molar mass of air, m₂ = molar mass of helium
340 / v = √4 / 28.8
v₂ = 340 / 0. 3727
v₂ = 912.26 m /s
fundamental frequency in helium = v₂ / 2L
fundamental frequency in helium = 912.26 / (2 x 0.625)
fundamental frequency in helium = 729.8 Hz
