Answer:
i think its changing shape if I'm wrong I'm sorry
Answer:
<u>When small organic molecules bind together, they form larger molecules called biological macromolecules.</u>Biological macromolecules are important cellular components and perform a wide array of functions necessary for the survival and growth of living organisms. The four major classes of biological macromolecules are carbohydrates, lipids, proteins, and nucleic acids.
(i hope this helps)
Answer:
40 g
Explanation:
Find the line labeled KClO3 (which might take you a min, theres a lot of lines here)
Notice that when the line creates a direct point, you can measure the exact temperature needed to dissolve a certain amount (like how they gave 30 degrees and it lined up perfectly with the 10 g line. )
Since its asking for the amount at 80 degrees, all you need to do is trace the line to the 80 degree point, and look at the grams. (notice it made a direct point, so there definitely should be any decimals or guesswork)
By reading the graph, you can tell that at 80 degrees, it dissolves 40 grams, and that is your answer.
Hope this helps :)
The force exerted on the parked car by the moving car is –30000 N
To solve this question, we'll begin by calculating the deceleration of the moving car. This can be obtained as follow:
Initial velocity (u) = 10 m/s
Final velocity (v) = 0 m/s
Time (t) = 0.5 s
<h3>Deceleration (a) =?</h3>
<h3>a = - 20 m/s</h3>
Finally, we shall determine the force exerted by the moving car on the parked car. This can be obtained as follow:
Mass of moving car (m) = 1500 Kg
Acceleration (a) = - 20 m/s
<h3>Force (F) =.?</h3>
F = ma
F = 1500 × (-20)
<h3>F = -30000 N</h3>
NOTE: The negative sign indicate force is in opposite direction to the parked car.
Therefore, the force exerted on the parked car by the moving car is –30000 N
Learn more: brainly.com/question/15430805
<span>Mole ratios are important to stoichiometric calculations because they bridge the gap when we have to convert between the mass of one substance and the mass of another.</span>
