Answer:
As there was no attached picture, I will explain how to take the measurement of liquids in any buret which you can then apply to the specific question
Explanation:
A buret is a laboratory apparatus used to precisely measure the volume of liquids (usually alkalise or bases) used in a titration experiment. The standard buret has a capacity of 50 ml and graduated in 0.1ml though burets with smaller capacities exist.
From the question, your buret is filled to the top (0.00ml) with liquid. It is very important when taking buret readings to place the buret below your eye level so that the bottom meniscus (lower part of the liquid) can be read.
To take the buret reading, note your initial buret reading (in this case 0.00ml) then titrate the liquid base in the buret against the acid by opening the tap located at the bottom of the buret.
When the titration or reaction is complete, note the final reading against the calibration of buret. You can do this by observing the lower meniscus of the liquid remaining in the buret. (Remember to keep the buret at eye level to avoid parallax error),
The difference between your final buret reading and the initial buret reading gives you the precise volume of liquid used in the reaction.
Answer:
B. to lock the tape into place
Explanation:
the button on the front of the housing locks the tape into place when pressed, preventing the tape from being pulled out further it retracting
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Answer:
the percent increase in the velocity of air is 25.65%
Explanation:
Hello!
The first thing we must consider to solve this problem is the continuity equation that states that the amount of mass flow that enters a system is the same as what should come out.
m1=m2
Now remember that mass flow is given by the product of density, cross-sectional area and velocity
(α1)(V1)(A1)=(α2)(V2)(A2)
where
α=density
V=velocity
A=area
Now we can assume that the input and output areas are equal
(α1)(V1)=(α2)(V2)
Now we can use the equation that defines the percentage of increase, in this case for speed
Now we use the equation obtained in the previous step, and replace values
the percent increase in the velocity of air is 25.65%
